If $(n + 2) .^{n} C_{0} . 2^{n + 1} - (n + 1)$ . $^{n} C_{1}$ . $2^{n} + n$ . $^{n} C_{2}$ . $2^{n - 1} +$ ... $= k (n + 1)$ , then the value of $k$ is

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Solution

Consider the expansion of $(x - 1)^{n}$ ,
$(x - 1)^{n}$ = $^{n} C_{0} x^{n} -^{n} C_{1} x^{n - 1} +^{n} C_{2} x^{n - 2} . . . +^{n} C_{n} (- 1)^{n}$
Multiplying by $x^{2}$
$x^{2} (x - 1)^{n} =^{n} C_{0} x^{n + 2} -^{n} C_{1} x^{n + 1} +^{n} C_{2} x^{n}$ .....
On differentiating we get,
$2 x (x - 1)^{n} + n x^{2} (x - 1)^{n - 1} = (n + 2) .^{n} C_{0} x^{(n + 1)} - (n + 1) .^{n} C_{1} x^{n} + . . . .$
Substitute $x = 2$ ,
$4 + 4 n = (n + 2) .^{n} C_{0} 2^{n + 1} - (n + 1) .^{n} C_{1} 2^{n} + n .^{n} C_{2} 2^{n - 1} +$ ...
$\Rightarrow 4 (n + 1) = k (n + 1)$
$\Rightarrow k = 4.$

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