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Question

(A) Given expression is $E = (1 + x)^n \left( 1 + \frac{1}{x} \right)^n$. We can rewrite this as $E = (1 + x)^n \left( \frac{x + 1}{x} \right)^n = \fr

Vedclass · Accepted Answer

$C_0^2 + C_1^2 + .... + C_n^2$

Vedclass · Answer

(A) Given expression is $E = (1 + x)^n \left( 1 + \frac{1}{x} \right)^n$. We can rewrite this as $E = (1 + x)^n \left( \frac{x + 1}{x} \right)^n = \frac{(1 + x)^{2n}}{x^n}$. The term independent of $x$ in the expansion of $E$ is the coefficient of $x^n$ in the expansion of $(1 + x)^{2n}$. The general term in the expansion of $(1 + x)^{2n}$ is given by $T_{r+1} = {}^{2n}C_r x^r$. To find the coefficient of $x^n$,we set $r = n$. Thus,the coefficient is ${}^{2n}C_n$. Using the identity ${}^{2n}C_n = \sum_{k=0}^{n} ({}^nC_k)^2$,we get the result as $C_0^2 + C_1^2 + .... + C_n^2$.

The term independent of $x$ in the expansion of ${(1 + x)^n}{\left( {1 + \frac{1}{x}} \right)^n}$ is

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