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

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Solution

$C_{1} + 2 C_{2} + 3 C_{3} + . . . . + n C_{n}$
$= n + \frac{2 n (n - 1)}{2!} + \frac{3 n (n - 1) (n - 2)}{3!} + . . . . . + n \times 1$
$= n + n (n - 1) + \frac{n (n - 1) (n - 2)}{2} + . . . . . . + n$
$= n [1 + n - 1 + \frac{(n - 1) (n - 2)}{2} + . . . . . + 1]$
$= n [1 + \frac{n - 1}{1!} + \frac{(n - 1) (n - 2)}{2} + . . . . . + 1]$
put $n - 1 = N$
$= n [1 + \frac{N}{1!} + \frac{N (N + 1)}{2!} + . . . . . + 1]$
$= n [N C_{0} + N C_{1} + N C_{2} + . . . . . . + N C_{N}]$
$= n 2^{N}$
$= n 2^{n - 1}$ Hence proved.

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