The expression $C_{0} + 2 C_{1} + 3 C_{2} + . . . . . . + (n + 1) C_{n}$ is equal to

2^{n - 1}

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n (2^{n - 1})

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n (2^{n - 1}) + 2^{n}

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(n + 1) 2^{n}

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Solution

The correct option is A $n (2^{n - 1}) + 2^{n}$
Let $E = C_{0} + 2 C_{1} + 3 C_{2} + . . . . . . n C_{n - 1} + (n + 1) C_{n} . . . . . . . . (1)$
Using $C_{r} = C_{n - r} E = (n + 1) C_{0} + n C_{1} + (n - 1) C_{2} + . . . . . + 2 C_{n - 1} + C_{n} . . . . . . . . (2)$
Adding $(1)$ and $(2)$ , we get
$2 E = (n + 2) C_{0} + (n + 2) C_{1} + (n + 2) C_{2} + . . . . + (n + 2) C_{n}$
$= (n + 2) {C_{0} + C_{1} + . . . . . + C_{n}} = (n + 2) 2^{n}$
$\Rightarrow E = (n + 2) 2^{n - 1}$

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