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

2^{n}

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n {.2}^{n}

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n {.2}^{n - 1}

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n {.2}^{n + 1}

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Solution

The correct option is B $n {.2}^{n - 1}$
$(1 + x)^{n} = \sum^{n} C_{r} x^{r}$
Differentiating with respect to $x$
$n x (1 + x)^{n - 1} = \sum^{n} C_{r} r x^{r - 1}$
$n x (1 + x)^{n - 1} =^{n} C_{1} + 2^{n} C_{2} x + 3^{n} C_{3} x^{2} + . . . n^{n} C_{n} x^{n - 1}$
Substituting $x = 1$ in the above expression gives us
$n 2^{n - 1} =^{n} C_{1} + 2^{n} C_{2} + 3^{n} C_{3} + . . . n^{n} C_{n}$ .

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Similar questions

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

C_{1} + 2 C_{2} + 3 C_{3} + 4 C_{4} + . +^{n} C_{n} = n {.2}^{n - 1}

C_{1} + 2 C_{2} + 3 C_{3} + . . n C_{n} = n {.2}^{n - 2}

I f (1 + x)^{n} = C_{0} + c_{1} x + . . . + C_{n} x^{n} +, t h e n \frac{C_{1}}{C_{0}} + \frac{2 C_{2}}{C_{1}} + \frac{3 C_{3}}{C_{2}} + . . . . . + \frac{n C_{n}}{C_{n - 1}} i s

C_{0}, C_{1}, C_{2}, . . . . . . . C_{n}

{(1 + x)}^{n}

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

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