CBSE Class 11 Maths Revision Notes Chapter 4

Q.1 Prove by the method of induction that every even power of every odd integer greater than 1 when divided by 8 leaves the remainder 1.

Ans

$$\begin{gathered} \mathrm{First}\hspace{0.17em}\mathrm{odd}\mathrm{integer}>1\mathrm{is}3, \\ \text{general Odd\hspace{0.17em}integer is\hspace{0.17em}}\left(2\mathrm{r}+1\right) \\ \text{To show} \\ {\left(2\mathrm{r}+1\right)}^{2\mathrm{n}}=8\mathrm{m}+1,\hspace{0.17em}\mathrm{m}\in \mathrm{N} \\ \text{i.e.\hspace{0.17em}}{\left(2\mathrm{r}+1\right)}^{2\mathrm{n}}–1\text{\hspace{0.17em}is divisible by\hspace{0.17em}}8 \\ \text{Let,\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right)\hspace{0.17em}:\hspace{0.17em}{\left(2\mathrm{x}+1\right)}^{2\mathrm{n}}–1=4{\mathrm{r}}^2+4\mathrm{r} \\ =4\mathrm{r}\left(\mathrm{r}+1\right) \\ \text{As\hspace{0.17em}}\mathrm{r}\left(\mathrm{r}+1\right)\text{is always even} \\ \therefore \hspace{0.17em}\hspace{0.17em}\mathrm{P}\left(1\right)\text{\hspace{0.17em}is true} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is true} \\ \mathrm{P}\left(\mathrm{k}\right)=2\mathrm{r}+1^{2\mathrm{k}}–1\text{\hspace{0.17em}is divisible by\hspace{0.17em}}8 \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true} \\ \text{i.e\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=2\mathrm{r}+{12}^{(\mathrm{k}+1)}–1\text{\hspace{0.17em}is divisible by\hspace{0.17em}}8 \\ \mathrm{P}\left(\mathrm{k}+1\right)={\left(2\mathrm{r}+1\right)}^{2\mathrm{k}+2}–1 \\ ={\left(2\mathrm{r}+1\right)}^{2\mathrm{k}}.{\left(2\mathrm{r}+1\right)}^2–1 \\ =\left(8\mathrm{m}+1\right)\left(8\mathrm{p}+1\right)–1 \\ =64\mathrm{mp}+8\mathrm{m}+8\mathrm{p}+1–1 \\ =8\left[8\mathrm{mp}+\mathrm{m}+\mathrm{p}\right]\text{\hspace{0.17em}Which is divisible by\hspace{0.17em}}8 \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is true} \\ \text{Hence by principle of mathematical induction}\mathrm{P}\left(\mathrm{n}\right)\text{is true . for all n\hspace{0.17em}∈\hspace{0.17em}N.} \end{gathered}$$

Q.2

$\mathbf{\text{Prove that}}{(1+X)}^{\mathbf{n}}\mathbf{\ge }(1+\mathrm{nx})\mathbf{\text{for all natural number}}\mathbf{n}\mathbf{,}\mathbf{\text{where}}\mathbf{x}\mathbf{>}\mathbf{–}\mathbf{1}$

Ans

$$\begin{gathered} Let\hspace{0.17em}\mathrm{P}\left(\mathrm{n}\right)=(1+\mathrm{x}{)}^{\mathrm{n}}\ge (1+\mathrm{nx}) \\ \Rightarrow \mathrm{P}\left(1\right)\text{is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)\hspace{0.17em}:\hspace{0.17em}(1+\mathrm{x}{)}^{\mathrm{k}}\ge (1+\mathrm{kx}) \\ \text{We Shall Prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is True for}\mathrm{x}>–1 \\ \text{Now}\mathrm{P}\left(\mathrm{k}+1\right)\hspace{0.17em}:\hspace{0.17em}(1+\mathrm{x}{)}^{\mathrm{k}+1} \\ (1+\mathrm{x}{)}^{\mathrm{k}+1}\ge \left(1+\mathrm{x}+\mathrm{kx}+{\mathrm{kx}}^2\right) \\ \text{Here}\mathrm{K}\text{is a natural number and}{\mathrm{x}}^2\ge 0\text{so that}{\mathrm{kx}}^2\ge 0 \\ \therefore \left(1+\mathrm{x}+\mathrm{kx}+{\mathrm{kx}}^2\right)\ge \left(1+\mathrm{x}+\mathrm{kx}\right) \\ \text{And so we obtain} \\ (1+\mathrm{x}{)}^{\mathrm{k}+1}\ge (1+\mathrm{x}+\mathrm{kx}\left) \\ \right(1+\mathrm{x}{)}^{\mathrm{k}+1}\ge [1+(1+\mathrm{k}\left)\mathrm{x}\right] \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right) \\ \text{Hence by principle of mathematical indention}\mathrm{P}\left(\mathrm{n}\right)\text{is true for where\hspace{0.17em}n\hspace{0.17em}∈\hspace{0.17em}N.} \end{gathered}$$

Q.3

$\mathbf{\text{Prove that}}{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{+}{\mathbf{n}}^{\mathbf{2}}\mathbf{>}\frac{{\mathbf{n}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{,}\mathbf{\hspace{0.17em}}\mathbf{n}\mathbf{\in }\mathbf{N}$

Ans

$$\begin{gathered} \text{Let}\mathrm{P}\left(\mathrm{n}\right)=1^2+2^2+\dots +{\mathrm{n}}^2>\frac{{\mathrm{n}}^3}{3} \\ \mathrm{P}\left(1\right)=1>\frac{1}{3} \\ \Rightarrow \mathrm{P}\left(1\right)\text{is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=1^2+2^2+\dots .+{\mathrm{k}}^2>\frac{{\mathrm{k}}^3}{3} \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{i.e}\mathrm{P}\left(\mathrm{k}+1\right)=1^2+2^2+\dots ..+{\mathrm{k}}^2+{\left(\mathrm{k}+1\right)}^2>\frac{{\left(\mathrm{k}+1\right)}^3}{3} \\ \text{Now}\mathrm{P}\left(\mathrm{k}+1\right)=\left(1+2^2+\dots .+{\mathrm{k}}^2\right)+{\left(\mathrm{k}+1\right)}^2>\frac{{\mathrm{k}}^3}{3}+{\left(\mathrm{k}+1\right)}^2 \\ =\frac{1}{3}\left[{\mathrm{k}}^3+3{\mathrm{k}}^2+6\mathrm{k}+3\right] \\ =\frac{1}{3}\left[(\mathrm{k}+1{)}^3+3\mathrm{k}+2\right]>\frac{1}{3}(\mathrm{k}+1{)}^3 \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{Hence by principle of mathematical induction is true. For all\hspace{0.17em}n\hspace{0.17em}}\in \mathrm{N}\text{.} \end{gathered}$$

Q.4 Prove that: 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24.

Ans

$$\begin{gathered} \text{Let\hspace{0.17em}P\hspace{0.17em}}\left(\mathrm{n}\right)=2.7^{\mathrm{n}}+3.5^{\mathrm{n}}–5 \\ \mathrm{P}\left(1\right)=2.7^1+3.5^1–5=14+15–5=24 \\ \Rightarrow \mathrm{P}\left(1\right)\text{\hspace{0.17em}is true} \\ \text{Assuming\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}\right)\text{\hspace{0.17em}is true} \\ \mathrm{P}\left(\mathrm{k}\right)=2.7^{\mathrm{k}}+3.5^{\mathrm{k}}–5\text{is divisible by}24 \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is true} \\ \text{i.e.\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=2.7^{\mathrm{k}+1}+3.5^{\mathrm{k}+1}–5\text{\hspace{0.17em}is divisible by\hspace{0.17em}}24 \\ 2.7^{\mathrm{k}+1}+3.5^{\mathrm{k}+1}–5=24\mathrm{g}\left(\mathrm{let}\right)...\left(1\right) \\ \mathrm{P}\left(\mathrm{k}+1\right)=2.7^{\mathrm{k}+1}+3.5^{\mathrm{k}+1}–5 \\ =2.7^{\mathrm{k}}.7+3.5^{\mathrm{k}}.5–5 \\ =7\left[2.7^{\mathrm{k}}+3.5^{\mathrm{k}}–5–3.5^{\mathrm{k}}+5\right]+3.5^{\mathrm{k}}.5–5 \\ =7\left[24\mathrm{g}–3.5^{\mathrm{k}}–5–3.5^{\mathrm{k}}–5\right]+15.5^{\mathrm{k}}–5\hspace{0.17em}\left[\text{From equation\hspace{0.17em}}\left(1\right)\right] \\ =7\times 24\mathrm{g}–21.5^{\mathrm{k}}+35+15.5^{\mathrm{k}}–5 \\ =7\times 24\mathrm{g}–6.5^{\mathrm{k}}+30 \\ =7\times 24\mathrm{g}–6\left(5^{\mathrm{K}}–5\right) \\ =7\times 24\mathrm{g}–6\left(4\mathrm{P}\right)\left[âˆ\mu \hspace{0.17em}{5}^{\mathrm{k}}–5\text{\hspace{0.17em}is a multiple of\hspace{0.17em}}{45}^{\mathrm{k}}–5=4\mathrm{P}\right] \\ =7\times 24\mathrm{g}–24\mathrm{P} \\ \text{Which is divisible by\hspace{0.17em}}24 \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is true} \\ \text{Hence by principle of mathematical induction is true. For all\hspace{0.17em}n ∈ N} \end{gathered}$$

Q.5

$$\begin{gathered} \mathbf{\text{Prove that:\hspace{0.17em}}}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{1}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{)}\mathbf{\text{\hspace{0.17em}…..\hspace{0.17em}}}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)} \\ \mathbf{\text{by using principle of mathematical induction.}} \end{gathered}$$

Ans

$$\begin{gathered} \text{Let}\mathrm{P}\left(\mathrm{n}\right)=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)⋯·\left(1+\frac{1}{\mathrm{n}}\right)=\left(\mathrm{n}+1\right) \\ \mathrm{P}\left(1\right)=\left(1+\frac{1}{1}\right)=\left(1+1\right) \\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2=2 \\ \Rightarrow \mathrm{P}\left(1\right)\text{is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\dots \left(1+\frac{1}{\mathrm{k}}\right)=\left(\mathrm{k}+1\right) \\ \text{We shall prove\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is True} \\ \text{i.e.,\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=\left(\mathrm{k}+2\right) \\ \text{Now\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)⋯\left(1+\frac{1}{\mathrm{k}}\right)\left(1+\frac{1}{\mathrm{k}+1}\right) \\ =\left(\mathrm{k}+1\right)\left(1+\frac{1}{\mathrm{k}+1}\right)=\left(\cancel{\mathrm{k}+1}\right)\frac{(\mathrm{k}+1+1)}{\left(\cancel{\mathrm{k}+1}\right)}=\left(\mathrm{k}+2\right) \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{Hence by mathematical induction}\mathrm{P}\left(\mathrm{n}\right)\text{is true for all\hspace{0.17em}}\mathrm{n}\in \mathrm{N}\text{.} \end{gathered}$$

Q.6 Prove that 7ⁿ – 3ⁿ is divisible by 4 for all n ∈ N.

Ans

$$\begin{gathered} Let\mathrm{P}\left(\mathrm{n}\right)=7^{\mathrm{n}}{3}^{\mathrm{n}} \\ \mathrm{P}\left(1\right)=7–3=4\text{which is divisible by}4 \\ \Rightarrow \mathrm{P}\left(1\right)\text{is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=7^{\mathrm{k}}–3^{\mathrm{k}}\text{is divisible by}4 \\ \text{Let}{7}^{\mathrm{k}}–3^{\mathrm{k}}=4\mathrm{d}...\left(1\right) \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{i.e.}\mathrm{P}\left(\mathrm{k}+1\right)=7^{\mathrm{k}+1}–3^{\mathrm{k}+1}\text{is divisible by}4 \\ \text{Now\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=7^{\mathrm{k}+1}–3^{\mathrm{k}+1}=7^{\mathrm{k}}.7–3^{\mathrm{k}}.3 \\ =7^{\mathrm{k}}.7–7.3^{\mathrm{k}}+7.3^{\mathrm{k}}–3.3^{\mathrm{k}} \\ =7\left(7^{\mathrm{k}}–3^{\mathrm{k}}\right)+\left(7–3\right).3^{\mathrm{k}} \\ =7\left(4\mathrm{d}\right)+4.3^{\mathrm{k}} \\ \text{Hence}\mathrm{P}\left(\mathrm{k}+1\right)\text{is divisible by}4 \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{Hence by principle of mathematical induction is true. For all\hspace{0.17em}n\hspace{0.17em}∈\hspace{0.17em}N} \end{gathered}$$

Q.7

$\mathbf{\text{Prove that\hspace{0.17em}}}\frac{\mathbf{1}}{\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{.}\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{.}\mathbf{4}}\mathbf{+}\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{3}\mathbf{)}}{\mathbf{4}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}}\mathbf{\text{\hspace{0.17em}for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}$

Ans

$$\begin{gathered} Let\hspace{0.17em}\mathrm{P}\left(\mathrm{n}\right)\hspace{0.17em}:\frac{1}{1.2·3}+\frac{1}{2.3.4}+\dots \dots +\frac{1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}=\frac{\mathrm{n}(\mathrm{n}+3)}{4(\mathrm{n}+1)(\mathrm{n}+2)} \\ \mathrm{P}\left(1\right)=\frac{1}{1.2.3}=\frac{1(1+3)}{4(1+1)(1+2)}\Rightarrow \frac{1}{6}=\frac{\cancel{4}}{\cancel{4}.2.3} \\ \text{⇒\hspace{0.17em}}\frac{1}{6}=\frac{1}{6}\Rightarrow \mathrm{P}\left(1\right)\text{\hspace{0.17em}is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\dots ..+\frac{1}{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)} \\ =\frac{\mathrm{k}(\mathrm{k}+3)}{4(\mathrm{k}+1)(\mathrm{k}+2)} \\ \text{We shall prove\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is True} \\ \text{i.e}\mathrm{P}\left(\mathrm{k}+1\right)=\frac{(\mathrm{k}+1)(\mathrm{k}+4)}{4(\mathrm{k}+2)(\mathrm{k}+3)} \\ \text{Now\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=\frac{\mathrm{k}(\mathrm{k}+3)}{4(\mathrm{k}+1)(\mathrm{k}+2)}+\frac{1}{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)} \\ =\frac{\mathrm{k}(\mathrm{k}+3{)}^2+4}{4(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)} \\ =\frac{\mathrm{k}\left({\mathrm{k}}^2+6\mathrm{k}+9\right)+4}{4(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)} \\ =\frac{{\mathrm{k}}^3+6{\mathrm{k}}^2+9\mathrm{k}+4}{4(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)} \\ =\frac{(\mathrm{k}+1{)}^2(\mathrm{k}+4)}{4(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+4)}=\frac{(\mathrm{k}+1)(\mathrm{k}+4)}{4(\mathrm{k}+2)(\mathrm{k}+3)} \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is True} \\ \text{Hence by principle of mathematical induction is true. For all\hspace{0.17em}}\mathrm{n}\in \mathrm{N} \end{gathered}$$

Q.8 Prove that 1.2 + 2.3 + 3.4 + … + n(n +1) = n( n +1)(n + 2)/3.

Ans

$$\begin{gathered} \text{Let\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right)\hspace{0.17em}\hspace{0.17em}:1.2+2.3+3.4+\dots +\mathrm{n}.\left(\mathrm{n}+1\right)=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3} \\ \mathrm{P}\left(1\right)=1.2=\frac{1(1+1)(1+2)}{3}\Rightarrow 2=\frac{2\times 3}{3}\Rightarrow 2=2 \\ \Rightarrow \mathrm{P}\left(1\right)\text{\hspace{0.17em}is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=1.2+2.3+3.4+\dots \dots +\mathrm{k}=\frac{\mathrm{k}(\mathrm{k}+1)\left(\mathrm{k}+2\right)}{3} \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is\hspace{0.17em}True} \\ \text{i.e\hspace{0.17em}}\mathrm{P}\left(\mathrm{k}+1\right)=1.2+2.3+3.4+\dots \dots +\mathrm{k}.\left(\mathrm{k}+1\right)+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right) \\ =\frac{(\mathrm{k}+1)\mathrm{k}+2\left)\right(\mathrm{k}+2)}{3} \\ \mathrm{P}\left(\mathrm{k}+1\right)=\frac{\mathrm{k}(\mathrm{k}+1)\mathrm{k}+2)}{3}+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right) \\ =\frac{\mathrm{k}(\mathrm{k}+1)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}{3} \\ =\frac{(\mathrm{k}+1)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}{3} \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is True} \end{gathered}$$

Q.9 Prove that : 2ⁿ > n for all positive integers n.

Ans

$$\begin{gathered} \text{When}\mathrm{n}=1, \\ {2}^1>1\text{. Hence}\mathrm{P}\left(1\right)\text{is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right):2^{\mathrm{k}}>\mathrm{k} \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{i.e.,}\mathrm{P}\left(\mathrm{k}+1\right)=2^{\mathrm{k}+1}>\mathrm{k}+1 \\ \mathrm{P}\left(\mathrm{k}+1\right)=2^{\mathrm{k}+1}=2^{\mathrm{k}}·2^1=2^{\mathrm{k}}·2 \\ \Rightarrow 2^{\mathrm{k}}.2>2\mathrm{k}=\mathrm{k}+\mathrm{k}>\mathrm{k}+1 \\ {2}^{\mathrm{k}+1}>2\mathrm{k}>\mathrm{k}+1 \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{Hence by principle induction}\mathrm{P}\left(\mathrm{n}\right)\text{is true, for all}\mathrm{n}\in \mathrm{N}\text{.} \end{gathered}$$

Q.10

$\mathbf{Prove}\mathbf{}\mathbf{that}\mathbf{}{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}{\mathbf{3}}^{\mathbf{3}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{n}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{6}}\mathbf{\text{\hspace{0.17em}for all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}$

Ans

$$\begin{gathered} \text{Let,\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right)=1^2+2^2+3^3+\dots +{\mathrm{n}}^2 \\ \mathrm{P}\left(1\right)=1^2=1^2\Rightarrow 1=1\Rightarrow \mathrm{P}\left(1\right)\text{is true} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=1^2+2^2+3^3+\dots \dots ..+{\mathrm{k}}^2=\frac{\mathrm{k}(\mathrm{k}+1)(2\mathrm{k}+1)}{6} \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is True} \\ \mathrm{P}\left(\mathrm{k}+1\right)=1^2+2^2+3^3+\dots \dots +{\mathrm{k}}^2+{\left(\mathrm{k}+1\right)}^2 \\ =\frac{(\mathrm{k}+1)(\mathrm{k}+2)(2\mathrm{k}+3)}{6} \\ \mathrm{P}\left(\mathrm{k}+1\right)=\frac{\mathrm{k}(\mathrm{k}+1)(2\mathrm{k}+1)}{6}+{\left(\mathrm{k}+1\right)}^2 \\ =\frac{\mathrm{k}\left(\mathrm{k}+1\right)\left(2\mathrm{k}+1\right)+6{\left(\mathrm{k}+1\right)}^2}{6} \\ =\frac{(\mathrm{k}+1)\left(2{\mathrm{k}}^2+\mathrm{k}+6\mathrm{k}+6\right)}{6} \\ =\frac{(\mathrm{k}+1)\left(2{\mathrm{k}}^2+7\mathrm{k}+6\right)}{6} \\ =\frac{(\mathrm{k}+1)\left(2{\mathrm{k}}^2+4\mathrm{k}+3\mathrm{k}+6\right)}{6} \\ =\frac{(\mathrm{k}+1)\left[2\mathrm{k}\right(\mathrm{k}+2)+3(\mathrm{k}+2\left)\right]}{6} \\ =\frac{(\mathrm{k}+1)(2\mathrm{k}+3)(\mathrm{k}+2)}{6}=\frac{(\mathrm{k}+1)(\mathrm{k}+2)(2\mathrm{k}+3)}{6} \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{\hspace{0.17em}is\hspace{0.17em}True} \\ \text{Hence, by principle of mathematical induction\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is\hspace{0.17em}true.\hspace{0.17em}For\hspace{0.17em}all\hspace{0.17em}}\mathrm{n}\in \mathrm{N} \end{gathered}$$

Q.11 Prove that: 1 + 3 + 5 +…+ (2n – 1) = n^2.

Ans

$$\begin{gathered} \text{Let,}\mathrm{P}\left(\mathrm{n}\right)=1+3+5+\dots ..+\left(2\mathrm{n}–1\right)={\mathrm{n}}^2 \\ \mathrm{P}\left(1\right)=1=1^2\Rightarrow 1=1\Rightarrow \mathrm{P}\left(1\right)\text{is True} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True} \\ \mathrm{P}\left(\mathrm{k}\right)=1+3+5+\dots \dots .+\left(2\mathrm{k}–1\right)={\mathrm{k}}^2 \\ \text{We shall prove}\mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{i.e.,}\mathrm{P}\left(\mathrm{k}+1\right)=1+3+5\dots \dots +\left(2\mathrm{k}–1\right)+\left(2\mathrm{k}+1\right)={\left(\mathrm{k}+1\right)}^2 \\ \text{Now}\mathrm{P}\left(\mathrm{k}+1\right)={\mathrm{k}}^2+2\mathrm{k}+1 \\ ={\left(\mathrm{k}+1\right)}^2 \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{Hence by principle of mathematical induction}\mathrm{P}\left(\mathrm{n}\right)\text{is true. For all}\mathrm{n}\in \mathrm{N} \end{gathered}$$

Q.12 Prove that: 1 + 2 + 3 +…+ n = n(n + 1)/2 for all n ∈ N.

Ans

$$\begin{gathered} \text{Let}\mathrm{P}\left(\mathrm{n}\right)=1+2+3+\dots \dots +\mathrm{n}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2} \\ \mathrm{P}\left(1\right)=1=\frac{1\left(1+1\right)}{2}\Rightarrow 1=1\Rightarrow \mathrm{P}\left(1\right)\text{is True.} \\ \text{Assuming}\mathrm{P}\left(\mathrm{k}\right)\text{is True.} \\ \mathrm{P}\left(\mathrm{k}\right)=1+2+3+\dots \dots +\mathrm{k}=\frac{\mathrm{k}\left(\mathrm{k}+1\right)}{2} \\ \text{Now}\mathrm{P}\left(\mathrm{k}+1\right)=1+2+3+\dots \dots +\mathrm{k}+\left(\mathrm{k}+1\right)=\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}{2} \\ =\frac{\mathrm{k}\left(\mathrm{k}+1\right)}{2}+\left(\mathrm{k}+1\right) \\ =\frac{{\mathrm{k}}^2+\mathrm{k}+2\mathrm{k}+2}{2} \\ =\frac{\mathrm{k}\left(\mathrm{k}+1\right)+2\left(\mathrm{k}+1\right)}{2} \\ =\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)}{2} \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is True} \\ \text{Hence by principle of mathematical induction}\mathrm{P}\left(\mathrm{n}\right)\text{is true for all}\mathrm{n}\in \mathrm{N}\text{.} \end{gathered}$$

Q.13 Suppose P(n): n(n+1)(n+2) is divisible by 6. Show that P(1), P(2) and P(3) are true.

Ans

Here
P(n) = n(n + 1)(n + 2)
P(1) = 1(1 + 1)(1 + 2) = 6 divisible by 6
P(2) = 2(2 + 1)(2 + 2) = 24 divisible by 6
P(1) = 3(3 + 1)(3 + 2) = 60 divisible by 6
Hence P(1), P(2) and P(3) are true.

Q.14 Let P(n)be statement 3n ≥ n!. where n is a natural number, then show that P(n) is true for n =1 and 2 .

Ans

P(1): 3 > 1! = 3 > 1 , which is true.
P(2): 6 > 2! = 6 > 2, which is true.

Q.15 If P(n) be the statement aⁿ + abⁿ is divisible by a, show that P(1) and P(2) are true.

Ans

P(n) : aⁿ + abⁿ is divisible by a
P(1) : a + b which is is divisible by a.
P(2): a² + ab² = a(a + b²), which is is divisible by a.
Hence P(1) and P(2) are true.

Q.16

$\mathbf{\text{If\hspace{0.17em}}}{\mathbf{A}}^{\mathbf{n}}\mathbf{=}\left[\begin{array}{cc}1& \text{na}\\ 0& 1\end{array}\right]\mathbf{,}\mathbf{\text{\hspace{0.17em}find\hspace{0.17em}}}\mathbf{A}\mathbf{\text{\hspace{0.17em}and\hspace{0.17em}}}{\mathbf{A}}^{\mathbf{2}}$

Ans

$$\begin{gathered} {\mathrm{A}}^{\mathrm{n}}=\left[\begin{array}{cc}1& \text{na}\\ 0& 1\end{array}\right], \\ \text{Substitute}\mathrm{n}=1,\text{we get} \\ \mathrm{A}=\left[\begin{array}{ll}1& \mathrm{a}\\ 0& 1\end{array}\right]\text{and} \\ {\mathrm{A}}^2=\left[\begin{array}{ll}1& \mathrm{a}\\ 0& 1\end{array}\right]\times \left[\begin{array}{ll}1& \mathrm{a}\\ 0& 1\end{array}\right]=\left[\begin{array}{ll}1& 2\mathrm{a}\\ 0& 1\end{array}\right]. \end{gathered}$$

Q.17 If P(n) : 7²ⁿ + 2^{3n – 3} 3^{n – 1} is divisible by 25, show that P(1) and P(2) are true.

Ans

P(n) : 7²ⁿ + 2^{3n – 3} 3^{n – 1}
P(1) : 7² + 2^{3 – 3} 3 ^{1 – 1} = 49 + 2⁰ 3 ⁰ = 49 + 1 = 50
P(2) : 7^2.2 + 2^{3.2 – 3} 3 ^{2 – 1} = 2401 + 24 = 2425
Which is divisible by 25 only.

Q.18 1³ + 2³ + 3³ + … + 1000³ = K² . Find the value of K.

Ans

We have,
1³ + 2³ + 3³ + … + 1000³ = K² , Σn³ = K² from n = 1 to 1000.
{1000(1000 + 1)/2}² = K²
Therefore, K = 250250.

Q.19 Show that a²ⁿ – b²ⁿ (a, b are distinct rational numbers) is divisible by a – b for n = 1, 2 and 3.

Ans

For n = 1,
a²ⁿ – b²ⁿ = a – b which is divisible by a – b
For n = 2,
a²ⁿ – b²ⁿ = a⁴ – b⁴ = (a² – b²)(a² + b²) = (a – b)(a + b)(a² + b²) Which is divisible by a – b
For n = 3
a²ⁿ – b²ⁿ = a⁶ – b⁶ = (a² – b²)(a² + b² + ab) = (a – b)(a + b)(a² + b² + ab) Which is divisible by a – b.

Q.20 What do you mean by Principle of mathematical induction?

Ans

To prove certain results or statements that are formulated in the terms of n with a well-suited principle that is user-based on the specific technique, where n is a positive integer, is known as the principle of mathematical induction.

Q.21 What is the main application of Principle of Mathematical Induction?

Ans

The principal of mathematical induction is used to prove the mathematical truth in the form of well defined statement or formula.

Q.22 What is the main drawback of Principle of Mathematical Induction?

Ans

The principal of mathematical Induction cannot be used to –

(i) establish a formula.
(ii) solve a equation.
(iii) calculate or operate any mathematical operation.
(iv) any operation other than proof.

Q.23 What do you mean by Principle of Mathematical Induction method?

Ans

In Principal of Mathematical Induction method, we proceed from particular cases to general cases.

Q.24 What do you mean by Principle of Mathematical Deduction method?

Ans

In Principle of Mathematical Deduction method, we proceed from general cases to particular cases.

Q.25 What is the basic assumption in Principle of Mathematical Induction?

Ans

First, we test P(n) for n = 1. If it is true, then we let n = k and on the basis of this, we test for n = k + 1.

Q.26 Write the algorithm for Principle of Mathematical Induction.

Ans

Step-I Obtain the statement P(n).
Step-II Test P(n) is true for n=1.
Step-III Let P(n) is true for n=k.
Step-IV Test P(n) is true for n=k+1 by using step III.

Q.27

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}{\mathbf{3}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{n}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{6}}\mathbf{\text{\hspace{0.17em}for all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N} \\ \mathbf{\text{then\hspace{0.17em}prove that\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{\text{\hspace{0.17em}is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(\mathrm{n}\right):1^2+2^2+3^2+\dots \dots ..+{\mathrm{n}}^2=\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6} \\ \mathrm{LHS}=1^2=1\left[\text{take first term, for}\mathrm{n}=1\right] \\ \text{Put}\mathrm{n}=1\text{in RHS} \\ \mathrm{RHS}=\frac{1\left(1+1\right)\left(2\times 1+1\right)}{6}=\frac{2\times 3}{6}=1 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(1\right)\text{is true} \end{gathered}$$

Q.28

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}{\mathbf{3}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{\dots }\mathbf{+}{\mathbf{n}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{6}}\mathbf{\text{for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N} \\ \mathbf{\text{then prove that\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{\text{\hspace{0.17em}is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(\mathrm{n}\right):1^2+2^2+3^2+\dots \dots +{\mathrm{n}}^2=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6} \\ \mathrm{LHS}=1^2+2^2=1+4=5\left[\text{take first two terms, for}\mathrm{n}=2\right] \\ \text{Put}\mathrm{n}=2\text{in RHS} \\ \mathrm{RHS}=\frac{2\left(2+1\right)\left(2\times 2+1\right)}{6}=\frac{2\times 3\times 5}{6}=5 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(2\right)\text{is true.} \end{gathered}$$

Q.29

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{+}\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{n}\mathbf{·}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\left[\frac{n(n+1)(n+2)}{3}\right] \\ \mathbf{\text{for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}\mathbf{,}\mathbf{\text{\hspace{0.17em}then prove that\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{\text{\hspace{0.17em}is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(\mathrm{n}\right):1.2+2.3+3.4+\dots ..+\mathrm{n}·\left(\mathrm{n}+1\right)=\left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}{3}\right] \\ \mathrm{LHS}=1.2=2\left[\text{take first term, for}\mathrm{n}=1\right] \\ \text{Put}\mathrm{n}=1\text{in}\mathrm{RHS} \\ \mathrm{RHS}=\frac{1\left(1+1\right)\left(1+2\right)}{3}=\frac{1\times 2\times 3}{3}=2 \\ \Rightarrow \text{LHS}=\text{RHS} \\ \Rightarrow \mathrm{P}\left(1\right)\text{is true.} \end{gathered}$$

Q.30

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{+}\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{n}\mathbf{·}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{\left[}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}}{\mathbf{3}}\mathbf{\right]} \\ \mathbf{\text{for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}\mathbf{,}\mathbf{\text{\hspace{0.17em}then prove that\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{\text{is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(\mathrm{n}\right):1.2+2.3+3.4+\dots ..+\mathrm{n}·\left(\mathrm{n}+1\right)=\left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}{3}\right] \\ \text{LHS}=1.2+2.3=2+6=8\left[\mathrm{take}\mathrm{first}\mathrm{two}\mathrm{term},\mathrm{for}\mathrm{n}=2\right] \\ \text{Put}\mathrm{n}=2\text{in RHS} \\ \mathrm{RHS}=\frac{2\left(2+1\right)\left(2+2\right)}{3}=\frac{2\times 3\times 4}{3}=8 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(2\right)\text{is true.} \end{gathered}$$

Q.31

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{1}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{)}\mathbf{\dots }\mathbf{·}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)} \\ \mathbf{\text{for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}\mathbf{,}\mathbf{\text{\hspace{0.17em}then prove that\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{\text{\hspace{0.17em}is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(\mathrm{n}\right):\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)⋯·\left(1+\frac{1}{\mathrm{n}}\right)=\left(\mathrm{n}+1\right) \\ \mathrm{LHS}=\left(1+\frac{1}{1}\right)=2\left[\text{take first term, for}\mathrm{n}=1\right] \\ \text{Put}\mathrm{n}=1\text{in RHS.} \\ \text{RHS}=\left(1+1\right)=2 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(1\right)\text{is true.} \end{gathered}$$

Q.32

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{1}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{)}\mathbf{}\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)} \\ \mathbf{\text{for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}\mathbf{\text{\hspace{0.17em}then, prove that}}\mathbf{P}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{\text{is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(\mathrm{n}\right):\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\dots ·\left(1+\frac{1}{\mathrm{n}}\right)=\left(\mathrm{n}+1\right) \\ LHS=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)=2\times \frac{3}{2}=3\left[\mathrm{take}\mathrm{first}\mathrm{two}\mathrm{terms},\mathrm{for}\mathrm{n}=2\right] \\ \text{Put}\mathrm{n}=2\text{in RHS} \\ \text{RHS}=(2+1)=3 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(2\right)\text{is true.} \end{gathered}$$

Q.33

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}\mathbf{7}\mathbf{+}\mathbf{77}\mathbf{+}\mathbf{777}\mathbf{+}\mathbf{7777}\mathbf{+}\mathbf{\dots }\mathbf{+}\mathbf{77}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\dots }\mathbf{\dots }\mathbf{.}\mathbf{7}\mathbf{=}\frac{\mathbf{7}}{\mathbf{81}}\mathbf{(}{\mathbf{10}}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{–}\mathbf{9}\mathbf{n}\mathbf{–}\mathbf{10}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{\text{n-digits}}\mathbf{\right)} \\ \mathbf{\text{for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}\mathbf{,}\mathbf{\text{\hspace{0.17em}then prove that}}\mathbf{P}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{\text{\hspace{0.17em}is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \text{If}\mathrm{P}\left(\mathrm{n}\right):7+77+777+7777+\dots +77\dots .......\dots .7=\frac{7}{81}\left({10}^{\mathrm{n}+1}–9\mathrm{n}–10\right) \\ \left(\text{n-digits}\right) \\ \text{for all}\mathrm{n}\in \mathrm{N} \\ \text{LHS}=7\left[\mathrm{take}\mathrm{first}\mathrm{term},\mathrm{for}\mathrm{n}=1\right] \\ \text{Put}\mathrm{n}=1\text{in RHS.} \\ \mathrm{RHS}=\frac{7}{81}\left({10}^{1+1}–9\times 1–10\right) \\ =\frac{7}{81}\left({10}^2–9–10\right) \\ =\frac{7}{81}\left(100–19\right) \\ =\frac{7}{81}\times 81 \\ =7 \\ \Rightarrow \mathrm{LHS}=\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(1\right)\text{is true.} \end{gathered}$$

Q.34

$$\begin{gathered} \mathbf{\text{If\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{}\mathbf{:}\mathbf{}\mathbf{7}\mathbf{+}\mathbf{77}\mathbf{+}\mathbf{777}\mathbf{+}\mathbf{7777}\mathbf{+}\mathbf{\dots }\mathbf{+}\mathbf{77}\mathbf{\dots }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{7}\mathbf{=}\frac{\mathbf{7}}{\mathbf{81}}\mathbf{(}{\mathbf{10}}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{–}\mathbf{9}\mathbf{n}\mathbf{–}\mathbf{10}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{\text{n-digits}}\mathbf{\right)} \\ \mathbf{\text{for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N}\mathbf{,}\mathbf{\text{then prove that\hspace{0.17em}}}\mathbf{P}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{\text{\hspace{0.17em}is true.}} \end{gathered}$$

Ans

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \text{If}\mathrm{P}\left(\mathrm{n}\right):7+77+777+7777+\dots +77\dots .......\dots .7=\frac{7}{81}\left({10}^{\mathrm{n}+1}–9\mathrm{n}–10\right) \\ \left(\text{n-digits}\right) \\ \text{for all}\mathrm{n}\in \mathrm{N} \\ \text{LHS}=7+77=84\left[\mathrm{take}\mathrm{first}\mathrm{term},\mathrm{for}\mathrm{n}=2\right] \\ \text{Put}\mathrm{n}=2\text{in RHS} \\ \mathrm{RHS}=\frac{7}{81}\left({10}^{2+1}–9\times 2–10\right) \\ =\frac{7}{81}\left({10}^3–18–10\right) \\ =\frac{7}{81}\left(1000–28\right) \\ =\frac{7}{81}\times 972 \\ =7\times 12=84 \\ \Rightarrow \text{LHS}=\text{RHS} \\ \Rightarrow \mathrm{P}\left(1\right)\text{is true} \end{gathered}$$

Q.35

$$\begin{gathered} \mathbf{\text{Using the principle of mathematical induction,}} \\ \mathbf{\text{prove that}} \\ \mathbf{1}\mathbf{.}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{+}\mathbf{\dots }\mathbf{+}\mathbf{n}(n+1)(n+2)\mathbf{=}\frac{\mathbf{n}(n+1)(n+2)(n+3)}{\mathbf{4}} \\ \mathbf{\text{for\hspace{0.17em}}}\mathbf{all}\mathbf{\hspace{0.17em}}\mathbf{n}\mathbf{\in }\mathbf{N} \end{gathered}$$

Ans

$$\begin{gathered} \text{Let the given statement be} \\ \mathrm{P}\left(\mathrm{n}\right)\hspace{0.17em}:\hspace{0.17em}1.2.3+2.3.4+\dots +\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)\left(\mathrm{n}+3\right)}{4}...\left(1\right) \\ \text{Test for\hspace{0.17em}}\mathrm{n}=1 \\ \mathrm{LHS}=1.2.3=6 \\ \mathrm{RHS}=\frac{1\times 2\times 3\times 4}{4}=6 \\ \text{Hence}\mathrm{P}\left(1\right)\text{\hspace{0.17em}is true.} \\ \text{Let\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for\hspace{0.17em}}\mathrm{n}=\mathrm{k},\text{\hspace{0.17em}then\hspace{0.17em}}\left(1\right)\text{\hspace{0.17em}becomes} \\ 1.2.3+2.3.4+\dots +\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)=\frac{\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}{4}...\left(2\right) \\ \text{Now, we shall test for\hspace{0.17em}}\mathrm{n}=\mathrm{k}+1\left(1\right)\text{\hspace{0.17em}becomes} \\ 1.2.3+2.3.4+\dots +\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)=\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)\left(\mathrm{k}+4\right)}{4}...\left(3\right) \\ \mathrm{LHS}=1.2.3+2.3.4+\dots +\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right) \\ =\frac{\mathrm{k}\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)}{4}+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)\hspace{0.17em}\mathrm{Using}\hspace{0.17em}\left(2\right) \\ =\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)\left[\frac{\mathrm{k}}{4}+1\right] \\ =\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right)\left(\mathrm{k}+4\right)}{4} \\ =\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is true.} \\ \text{Hence,\hspace{0.17em}}\mathrm{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for all\hspace{0.17em}}\mathrm{n}\in \mathrm{N}\text{.} \end{gathered}$$

Q.36

$$\begin{gathered} \mathbf{\text{Using the principle of mathematical induction,}} \\ \mathbf{\text{prove that}} \\ \mathbf{1}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\dots }\mathbf{+}\mathbf{n}\mathbf{.}(n+2)\mathbf{=}\frac{\mathbf{n}(n+1)(2n+7)}{\mathbf{6}} \\ \mathbf{\text{for\hspace{0.17em}all\hspace{0.17em}}}\mathbf{n}\mathbf{\in }\mathbf{N} \end{gathered}$$

Ans

$$\begin{gathered} \text{Let the given statement be} \\ \mathrm{P}\left(\mathrm{n}\right)\hspace{0.17em}:\hspace{0.17em}1.3+2.4+3.5\dots +\mathrm{n}.\left(\mathrm{n}+2\right)=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+7\right)}{6}...\left(1\right) \\ \text{Test for\hspace{0.17em}}\mathrm{n}=1 \\ \text{LHS\hspace{0.17em}}=1.3=3 \\ \mathrm{RHS}=\frac{1\times 2\times 9}{6}=3 \\ \text{Hence,}\mathrm{P}\left(1\right)\text{is true.} \\ \text{Let}\mathrm{P}\left(\mathrm{n}\right)\text{is true for}\mathrm{n}=\mathrm{k}\text{then}\left(1\right)\text{becomes} \\ 1.3+2.4+3.5\dots +\mathrm{k}\left(\mathrm{k}+2\right)=\frac{\mathrm{k}\left(\mathrm{k}+1\right)\left(2\mathrm{k}+7\right)}{6}...\left(2\right) \\ \text{Now, we shall test for}\mathrm{n}=\mathrm{k}+1\left(1\right)\text{becomes} \\ 1.3+2.4+3.5\dots \mathrm{k}\left(\mathrm{k}+2\right)+\left(\mathrm{k}+1\right)\left(\mathrm{k}+3\right)=\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(2\mathrm{k}+9\right)}{6}...\left(3\right) \\ \text{LHS}=1.3+2.4+3.5\dots \mathrm{k}\left(\mathrm{k}+2\right)+\left(\mathrm{k}+1\right)\left(\mathrm{k}+3\right) \\ =\frac{\mathrm{k}\left(\mathrm{k}+1\right)\left(2\mathrm{k}+7\right)}{6}+\left(\mathrm{k}+1\right)\left(\mathrm{k}+3\right)\hspace{0.17em}\mathrm{Using}\hspace{0.17em}\left(2\right) \\ =\left(\mathrm{k}+1\right)\left[\frac{\mathrm{k}(2\mathrm{k}+7)}{6}+\mathrm{k}+3\right] \\ =\frac{\left(\mathrm{k}+1\right)\left[2{\mathrm{k}}^2+13\mathrm{k}+18\right]}{6} \\ =\frac{\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)\left(2\mathrm{k}+9\right)}{6} \\ =\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is true} \\ \text{Hence,}\mathrm{P}\left(\mathrm{n}\right)\text{is true for all}\mathrm{n}\in \mathrm{N}\text{.} \end{gathered}$$

Q.37

$$\begin{gathered} \mathbf{\text{Using the principle of mathematical induction,}} \\ \mathbf{\text{prove that}} \\ \mathbf{1}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{\dots }\mathbf{+}(2n–1)(2n+1)\mathbf{=}\frac{\mathbf{n}(4n^2+6n–1)}{\mathbf{3}} \\ \mathbf{\text{for\hspace{0.17em}}}\mathbf{all}\mathbf{\hspace{0.17em}}\mathbf{n}\mathbf{\in }\mathbf{N}\mathbf{.} \end{gathered}$$

Ans

$$\begin{gathered} \text{Let the given statement be} \\ \mathrm{P}\left(\mathrm{n}\right)\hspace{0.17em}:\hspace{0.17em}1.3+3.5+5.7\dots +\left(2\mathrm{n}–1\right).\left(2\mathrm{n}+1\right)=\frac{\mathrm{n}\left(4{\mathrm{n}}^2+6\mathrm{n}–1\right)}{3}...\left(1\right) \\ \text{Test for\hspace{0.17em}}\mathrm{n}=1 \\ \mathrm{LHS}=1.3=3 \\ \mathrm{RHS}=\frac{1\times (4+6–1)}{3}=\frac{9}{3}=3 \\ \text{Hence,}\mathrm{P}\left(1\right)\text{is true.} \\ \text{Let}\mathrm{P}\left(\mathrm{n}\right)\text{is true for}\mathrm{n}=\mathrm{k}\text{then}\left(1\right)\text{becomes} \\ 1.3+3.5+5.7\dots +\left(2\mathrm{k}–1\right).\left(2\mathrm{k}+1\right)=\frac{\mathrm{k}\left(4{\mathrm{k}}^2+6\mathrm{k}–1\right)}{3}...\left(2\right) \\ \text{Now, we shall test for}\mathrm{n}=\mathrm{k}+1\hspace{0.17em}\left(1\right)\text{\hspace{0.17em}becomes} \\ 1.3+3.5+\dots +\left(2\mathrm{k}+1\right)\left(2\mathrm{k}+3\right)=\frac{\left(\mathrm{k}+1\right)\left(4{\mathrm{k}}^2+14\mathrm{k}+9\right)}{3}...\left(3\right) \\ \mathrm{LHS}=1.3+3·5+\dots +\left(2\mathrm{k}–1\right).\left(2\mathrm{k}+1\right)+\left(2\mathrm{k}+1\right)\left(2\mathrm{k}+3\right) \\ =\frac{\mathrm{k}\left(4{\mathrm{k}}^2+6\mathrm{k}–1\right)}{3}+\left(2\mathrm{k}+1\right)\left(2\mathrm{k}+3\right)\text{\hspace{0.17em}Using\hspace{0.17em}}\left(2\right) \\ =\left[\frac{4{\mathrm{k}}^3+6{\mathrm{k}}^2–\mathrm{k}}{3}+\left(4{\mathrm{k}}^2+8\mathrm{k}+3\right)\right] \\ =\frac{4{\mathrm{k}}^3+6{\mathrm{k}}^2–\mathrm{k}+12{\mathrm{k}}^2+24\mathrm{k}+9}{3} \\ =\frac{4{\mathrm{k}}^3+18{\mathrm{k}}^2+23\mathrm{k}+9}{3} \\ =\frac{(\mathrm{k}+1)\left(4{\mathrm{k}}^2+14\mathrm{k}+9\right)}{3}=\mathrm{RHS} \\ \Rightarrow \mathrm{P}\left(\mathrm{k}+1\right)\text{is true.} \\ \text{Hence,}\mathrm{P}\left(\mathrm{n}\right)\text{is true for all}\mathrm{n}\in \mathrm{N}\text{.} \end{gathered}$$