If ${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . C_{n} x^{n}$ , then prove that $C_{1} + {2 C}_{2} + 3 C_{3} . . . . {+ n C}_{n} =$

n . 2^{n}

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n . 2^{n + 1}

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n . 2^{n - 1}

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n + 1. 2^{n - 1}

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Solution

The correct option is B $n . 2^{n - 1}$
Here the last term of $C_{1} + C_{2} + 3 C_{3} + . . . + n C_{n} i s n C_{n}$ i.e., $n$ and last term with positive sign. Then $n = n .1 + 0$ or $n \div n = 1$ Here $q = 1$ and $r = 0$ and the given series is ${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . C_{n} x^{n}$ then differentiating both sides w.r.t to $x$ , we get
$\Rightarrow n {(1 + x)}^{n} = C_{0} + C_{1} + {2 C}_{2} x + 3 C_{3} x^{2} . . . . + n C_{n} x^{n - 1}$
Putting x=1, we get
$n . 2^{n - 1} = C_{1} + {2 C}_{2} x + 3 C_{3} x^{2} . . . . + n C_{n} o r C_{1} + {2 C}_{2} x + 3 C_{3} x^{2} . . . . + n C_{n} = n . 2^{n - 1}$

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