$C_{0} + 2 C_{1} + 3 C_{2} + 4 C_{3} + . . . + (n + 1) C_{n} = (n + 2) 2^{n - 1}$

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Solution

Consider the following question.

$C_{0} + 2 C_{1} + 3 C_{2} + 4 C_{3} + . . . + (n + 1) C_{n} = (n + 2) 2^{n - 1}$

${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n} . . . . . (1)$

Multiplying $(1)$ with $x$ , we get.

$x {(1 + x)}^{n} = C_{0} x + C_{1} x^{2} + C_{2} x^{3} + . . . + C_{n} x^{n + 1} . . . . . . (2)$

${(1 + x)}^{n} + n {(1 + x)}^{n 1} x = C_{0} x + 2 C_{1} x^{2} + . . . + (n + 1) C_{n} x^{n} . . . (3)$

Putting $= 1 i n (3),$ we text get.

$2 n + n {.2}^{n 1} = C_{0} + 2 C_{1} + 3 C_{2} + . . . + (n + 1) C_{n}$

$C_{0} + 2 C_{1} + 3 C_{2} + . . . + (n + 1) C_{n} = 2^{n - 1} (n + 2)$

Hence, this is the required answer.

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