The value of $(n + 2) C_{0} 2^{n + 1} - (n + 1) C_{1} 2^{n} + n C_{2} 2^{n - 1} + . . .$ is

4

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4 n

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4 (n + 1)

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2 (n + 2)

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Solution

The correct option is A $4 (n + 1)$
Let $n = 2$ , we get
$4^{2} C_{0} 2^{3} - (3)^{2} C_{1} 2^{2} + 2^{2} C_{2} 2^{1}$
$= 4 (8) - 3 (2.4) + 2 (2)$
$= 32 - 24 + 4$
$= 36 - 24$
$= 12$
$= 4. (3)$
$= 4 (1 + 2)$
$= 4 (1 + n)$ .

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Similar questions

(n + 2) C_{0} 2^{n + 1} - (n + 1) C_{1} 2^{n} + n C_{2} 2^{n - 1} + . . . .

(C_{r} =^{n} C_{r})

2^{n} - (n - 1) 2^{n - 2} + \frac{(n - 2) (n - 3)}{⌊ 2} - \frac{(n - 3) (n - 4) (n - 5)}{⌊ 3} 2^{n - 6} + . . . . .;

3

1 - (n - 1) + \frac{(n - 2) (n - 2)}{⌊ 2} - \frac{(n - 3) (n - 4) (n - 5)}{⌊ 3} + . . . . = (- 1)^{n}

\frac{2 n!}{(n!)^{2}} > \frac{4 n}{2 n + 1}

2 n

a^{'} s

b^{'} s,

lim n \to \infty (\frac{1}{\sqrt{4 n^{2} - 1}} + \frac{1}{\sqrt{4 n^{2} - 4}} + . . . . + \frac{1}{\sqrt{4 n^{2} - 2 n}})

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