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JEE Mains · Maths · STD 11 - 7. binomial theoram

\((1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5, x \neq 0 \) के प्रसार में \(\mathrm{x}^3\) तथा \(\mathrm{x}^{-13}\) के गुणांकों का योग ........... है।

  1. A \(118\)
  2. B \(116\)
  3. C \(115\)
  4. D \(117\)
Verified Solution

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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