Question 9 Exercise 7.3
Solutions of Question 9 of Exercise 7.3 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Q9 Find the coefficient of $x^{\prime \prime}$ in $\left(\frac{1+x}{1-x}\right)^2$. Solution: Given that: $$ \begin{aligned} & \left(\frac{1+x}{1-x}\right)^2=(1+x)^2(1-x)^{-2} \\ & =\left(x^2+2 x+1\right)(1-x)^2 \end{aligned} $$
Applying binomial theorem $$ \begin{aligned} & =\left(x^2+2 x+1\right)[1+2 x+ \\ & \frac{-2(-2-1)}{2 !}(-x)^2 \\ & \left.+\frac{\cdots 2(-2-1)(-2-2)}{3 !}(-x)^3+\cdots\right] \\ & =\left(i^2+2 x+1\right)\left[1+2 x+3 x^2+4 x^3\right. \\ & +\ldots . .] \end{aligned} $$ Generalizing up-to $x^{\prime t}$ as $$ \begin{aligned} & =\left(x^2+2 x+1\right)\left[1+2 x+3 x^3+4 x^3+\right. \\ & \ldots+(n-2) x^{n-3}+(n-1) x^{n-2}+n x^{n-1} \\ & \left.+(n+1) x^n+\ldots .\right] \end{aligned} $$
Multiplying and just collecting the terms containing $x^n$ $$ \begin{aligned} & (n+1) x^n+2 n x^n+(n-1) x^n \\ & =(n+1+2 n+n-1) x^n \\ & =4 n x^n . \end{aligned} $$
Hence the coelficient of $x^n$ in $\left(\frac{1+x}{1-x}\right)^2$ is $4 n$.
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