Binomial Theorem
A binomial expression has two terms, which are connected by the operators ‘+’ or ‘–’. The total number of terms in the expansion of (a + b)n, for any positive integral n is (n + 1), i.e., one more than the index. In each term of the expansion of (a + b)n, for any positive integral n, the sum of the indices of a and b is same and equal to n. For any positive integer n, the binomial theorem is given by (a + b)n = nc0anb0 + nc1an – 1b1 + nc2an – 2b2 + … + ncna0bn, where nN. In the expansion of (a + b)n , nc0, nc1, nc2, ncn are called the binomial coefficients. The general term in the expansion of (a + b)n is given by (r + 1)th term, i.e., Tr + 1 = ncran – rbr, where 0 ≤ r ≤ n. In case of an even index, the number of the middle terms in the expansion of (a + b)n is one and is given by [(n/2) + 1]th term. In case of an odd index, the number of the middle terms in the expansion of (a + b)n is two which are given by [(n + 1)/2]th term and [(n + 1)/2 + 1]th term.
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Q1
5^{th} term in the expansion of (x  y)^{6} is
Marks:1Answer:
15 x^{2} y^{4}.
Explanation:
The general term in the expansion of (x  y)^{6 }is given by
T_{r}_{+1 }= ^{6}C_{r}(x)^{6r}.(y)^{r}T_{4+1 }= (1)^{4}.^{6}C_{4}.x^{64}y^{4}
=15 x^{2} y^{4}

Q2
The position of middle term in the expansion of (1 + x)^{20} is
Marks:1Answer:
11^{th} term.
Explanation:
Total number of terms = 20+1 =21
Middle term = (21+1)/2 = 11^{th} term

Q3
The position of middle term in the expansion of (1 + x)^{20} is
Marks:1Answer:
11^{th} term.
Explanation:
Total number of terms = 20+1 =21
Middle term = (21+1)/2 = 11^{th} term

Q4
5^{th} term in the expansion of (x  y)^{6} is
Marks:1Answer:
15 x^{2} y^{4}.
Explanation:
The general term in the expansion of (x  y)^{6 }is given by
T_{r}_{+1 }= ^{6}C_{r}(x)^{6r}.(y)^{r}T_{4+1 }= (1)^{4}.^{6}C_{4}.x^{64}y^{4}
=15 x^{2} y^{4}

Q5Marks:1
Answer:
r + 1.
Explanation: