Given that $C_{1} + 2 C_{2} x + 3 C_{3} x^{2} + . . . + 2 n C_{2 n} x^{2 n - 1} = 2 n (1 + x)^{2 n - 1}$ , where $C_{r} = (2 n)! / [r! (2 n - r)!]; r = 0, 1, 2, . . . ., 2 n$ , then prove that $C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - . . . - 2 n C_{2 n}^{2} = (- 1)^{n} n C_{n}$ .

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Solution

Given that
$C_{1} + 2 C_{2} + 3 C_{3} x^{2} + . . . + 2 n C_{2 n} x^{2 n - 1} = 2 n (1 + x)^{2 n - 1}$
where
$C_{r} = \frac{2 n!}{r! (2 n - r)!}$
Integrating both sides with respect to $x$ , under the limits $0$ to $x$
we get
$[C_{1} x + C_{2} x^{2} + C_{3} x^{3} + . . . . + C_{2 n} x^{2 n}]_{0}^{x} [(1 + x)^{2 n}]_{0}^{x}$
$\Rightarrow C_{1} x + C_{2} x^{2} + C_{3} x^{3} + . . . . + C_{2 n} x^{2 n} = (1 + x)^{2 n} - 1$
$\Rightarrow C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} + . . . . + C_{2 n} x^{2 n} = (1 + x)^{2 n}$
Changing $x$ by $- 1 / x$ , we get
$C_{0} - \frac{C_{1}}{x} + \frac{C_{2}}{x^{2}} - \frac{C_{3}}{x^{3}} + . . . . + (- 1)^{2 n} \frac{C_{2 n}}{x^{2 n}} = {(1 - \frac{1}{x})}^{2 n}$
$\Rightarrow C_{0} n^{2 n} - C_{1} x^{2 n - 1} + C_{2} x^{2 n - 2} + C_{3} x^{2 n - 3} + . . . + C_{2 n} = (x - 1)^{2 n}$
Multiplying equation (3) and equating the coefficient of $x^{2 n - 1}$ on both sides, we get
$- C_{1}^{2} + 2 C_{2}^{2} - 3 C_{3}^{2} + . . . + 2 n C_{2 n}^{2}$
= coefficient of $x^{2 n - 1}$ in $(x^{2} - 1)^{2 n - 1} (x - 1)$
$= 2 n$ [coefficient of $x^{2 n - 2}$ in $(x^{2} - 1)^{2 n - 1} -$ coefficient of $x^{2 n - 1}$ in $(x^{2} -)^{2 n - 1}$ ]
$= 2 n [^{2 n - 1} C_{n - 1} (- 1)^{n - 1} - 0]$
$= (- 1)^{n - 1} 2 n^{2 n - 1} C_{n - 1}$
$\Rightarrow C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} + . . . . + 2 n C_{2 n}^{2}$
$= (- 1)^{n} 2 n^{2 n - 1} C_{n - 1}$
$= (- 1)^{n} n \times (\frac{2 n}{n} \times^{2 n - 1} C_{n - 1})$
$= (- 1)^{n} n^{2 n} C_{n}$
$= (- 1)^{n} n C_{n} (∵^{2 n} C_{n} = C_{n})$

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