If $(1 + x)^{n} = c_{0} + c_{1} x + c_{2} x^{2} + . . . . + c_{n} x^{n}$ , find the value of $1^{2} c_{1} + 2^{2} c_{2} + 3^{2} c_{3} + . . . . . + n^{2} c_{n}$ .

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Solution

we can easily show that
$1^{2} + 2^{2} x + 3^{2} x^{2} + 4^{2} x^{3} + . . . + n^{2} x^{n - 1} + . . . . . = \frac{1 + x}{(1 - x)^{3}}$ .
Also $c_{n} + c_{n - 1} x + . . . . c_{2} x^{n - 2} + c_{1} x^{n - 1} + c_{0} x^{n} = (1 + x)^{n}$ .
Multiply together these two results $;$ then the given series is equal to the coefficient of $x^{n - 1}$ in $\frac{(1 + x)^{n + 1}}{(1 - x)^{3}},$ that is, in $\frac{(2 - ¯ ¯¯¯¯¯¯¯¯¯¯¯ ¯ 1 - x)^{n + 1}}{(1 - x)^{3}}$ .
The only terms containing $x^{n - 1}$ in this expansion arise from
$2^{n + 1} (1 - x)^{- 3} - (n + 1) 2^{n} (1 - x)^{- 2} + \frac{(n + 1) n}{⌊ 2} 2^{n - 1} (1 - x)^{- 1};$
$∴$ the given series $= \frac{n (n + 1)}{⌊ 2} 2^{n + 1} - n (n + 1) 2^{n} + \frac{n (n + 1)}{⌊ 2} 2^{n - 1}$
$= n (n + 1) 2^{n - 2}$

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