In the usual notations prove that
$C_{1} + 2 C_{2} x + 3 C_{3} x^{2} + . . . . . n C_{n} x^{n - 1} = n (1 + x) n - 1$

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Solution

Alf. Method :
$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} . . . . . . . + C_{n} x^{n} .$
Differentiate both sides w.r.t. x.
$n (1 + x)^{n - 1} = C_{1} + 2 C_{2} x + 3 C_{3} x^{2} . . . . . . . + n C_{n} x^{n - 1} .$ .......(1)
This prove the result.
Putting x = 1, we get
$n \cdot 2^{n - 1} = C_{1} + 2 C_{2} + 3 C_{3} . . . . . . . + n C_{n}$
This is Q .1 (a) done above
Putting x = -1 we get
$0 = C_{1} - 2 C_{2} + 3 C_{3} - . . . . .$
This is Q .1 (b) done above.

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