Prove that :
$C_{1} + 2 C_{2} + 3 C_{3} + . . n C_{n} = n {.2}^{n - 2}$

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Solution

$C_{1} + 2 C_{2} + 3 C_{3} + . . . . . + n C_{n}$
$= n + 2 \cdot \frac{n (n - 1)}{2!} + 3 \cdot \frac{n (n - 1) (n - 2)}{3!} + . . . . + n \cdot 1$
$= n [1 + \frac{n - 1}{1!} + \frac{(n - 1) \cdot (n - 2)}{2!} + . . . . 1]$
Put $n - 1$ = N.
$∴$ R.H.S. = $n [1 + N + \frac{N (N - 1)}{2!} + . . . . . .1]$
$= n [^{N} C_{0} +^{N} C_{1} +^{N} C_{2} + . . . . . +^{N} C_{N}]$
$= n 2^{N} = n \cdot 2^{n - 1}$

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