$C_{1} + 2 C_{2} + 3 C_{3} + 4 C_{4} + . . . . . . . . . + n + 1 n C_{n}$

n 2^{n - 1}

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2^{n + 1}

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{2.2}^{n - 1}

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Zero

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Solution

The correct option is A $n 2^{n - 1}$
$\begin{matrix} C_{1} + 2 C_{2} + 3 C_{3} + . . . . . . + n C_{n} = n + 2 \times \frac{n (n - 1)}{2!} + 3 \times \frac{n (n - 1) (n - 2)}{3!} + . . . . + n \times 1 = n + 2 \times \frac{n (n - 1)}{1} + 3 \times \frac{n (n - 1)}{6} + . . . . + n = n + \frac{n (n - 1)}{1} + \frac{n (n - 1)}{2} + . . . . . + n = n [1 + \frac{n - 1}{1} + \frac{n (n - 1)}{2} + . . . . . + 1] = n [1 + \frac{n - 1}{1!} + \frac{n (n - 1)}{2!} + . . . . . + 1] p u t n - 1 = N = n [1 + \frac{N}{1!} + \frac{N (N + 1)}{2!} + . . . . . + 1] = n (^{N} C_{0} +^{N} C_{1} +^{N} C_{2} + . . . . +^{N} C_{N}) = n 2^{N} = n 2^{n - 1} H e n c e, t h e o p t i o n A i s t h e c o r r e c t a n s w e r \end{matrix}$

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