Prove that according as n is odd or even the value of $C_{0}^{2} - C_{1}^{2} + C_{2}^{2} - . . . . . . + (- 1)^{n} (C_{n})^{2} = 0$
or $(- 1)^{n / 2} \frac{n!}{(n / 2)! (n / 2)!}$

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Solution

$(1 + x)^{n} = C_{0} + C_{1} x C_{2} x^{2} + . . . . + C_{n} x^{n}$
$(x - 1)^{n} = C_{0} x^{n} - C_{1} x^{x - 1} + C_{2} x^{x - 2} + . . . . + (- 1)^{n} C_{n}$
Multiplying both the sides
$(- 1)^{n} (1 - x^{2})^{n} = () ()$
Now $C_{0}^{2} - C_{1}^{2} + C_{2}^{2} -$ is the coefficient of $x^{n}$ in the product in R. H. S. . Hence it is the coefficient of $x^{n} i n (- 1)^{n} (1 - x^{2})^{2}$ of coeff of $(x^{2})^{n / 2} i n (1 - x^{2})^{n}$ Which will appear in $T_{\frac{n}{2} + 1}$
$∴ (- 1)^{n}^{n} C_{n / 2} (- 1)^{n / 2} (x^{2})^{n / 2}$
Above is possible only when $\frac{n}{2}$ is an integer . i.e. n is even and is case n is odd , then the term $x^{n}$ will not occur .Also when n is even , then $(- 1)^{n} = 1$
$∴ (- 1)^{n / 2} \cdot \frac{n!}{\frac{n}{2}! \frac{n}{2}!}$ is the required answer

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