If $(1 + x)^{n} = c_{0} + c_{1} x + c_{2} x^{2} + c_{3} x^{3} + . . . . + c_{n} x^{n},$ n being a positive integer, find the value of $(n - 1)^{2} c_{1} + (n - 3)^{2} c_{3} + (n - 5)^{2} c_{5} + . . . . .$

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Solution

$\Rightarrow {(e^{x} + 1)}^{n} = e^{n x} + c_{1} e^{(n - 1) x} + . . . . + c_{n} e^{0 n}$
${(e^{x} - 1)}^{n} = e^{n x} - c_{1} e^{(n - 1) x} + . . . . + c_{n} e^{0 x} {(- 1)}^{n}$
$2 {c_{1} e^{(n - 1) x} + c_{3} e^{(n - 3) x} + . . . .} = {(e^{x} + 1)}^{n} - {(e^{x} - 1)}^{n}$
Equating co efficients of $x^{2}$ we have,
$\frac{2 S}{2!} =$ coefficient of $x^{2}$ in ${(e^{x} + 1)}^{n} - {(e^{x} - 1)}^{n}$
i.e., in ${(2 + x + \frac{x^{2}}{2})}^{n}$ or in $2^{n - 2} n (x + \frac{x^{2}}{2}) + \frac{n (n - 1)}{2!} 2^{n - 2} x^{2}$
$∴ S = n . 2^{n - 2} + n (n - 1) 2^{n - 3}$
$= n (n + 1) 2^{n - 3}$

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