JEE Mains · Maths · STD 11 - 7. binomial theoram
In the expansion of \((1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5, x \neq 0\), the sum of the coefficient of \(x^3\) and \(x^{-13}\) is equal to
- A \(118\)
- B \(116\)
- C \(115\)
- D \(117\)
Answer & Solution
Correct Answer
(A) \(118\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & (1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5 \\ & =(1+x)\left(1-x^2\right)\left(\left(1+\frac{1}{x}\right)^3\right)^5 \\ & =\frac{(1+x)^2(1-x)(1+x)^{15}}{x^{15}} \\ & =\frac{(1+x)^{17}-x(1+x)^{17}}{x^{15}} \\ &…
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