====== 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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