Question

If ${(1+x)}^n=\sum _{r=0}^n{C}_r{x}^r=C_0+C_{1}x+C_2{x}^2+...+C_n{x}^n$ , then calculated value of $C_0^2-C_1^2+C_2^2-C_3^2+...(-1{)}^n{C}_n^2$ if $n=15$ is equals

• A. $\frac{2n!}{15!14!}$
• B. $\frac{15!}{\frac{15}{2}!\frac{15}{2}!}(-1)$
• C. $\frac{2n!}{15!15!}$
• D. $0$

Answer 1

The correct option is D $0$

Method (i):

Let S $=C_0^2-C_1^2+C_2^2-C_3^2+....$
Given ${(1+x)}^n=C_0{x}^n+C_n{x}^{n-1}+\cdots +C_n{x}^0\cdots \left(A\right)$ and ${(1+x)}^n=C_0-C_{1}x+C_2{x}^2+\cdots +(-1^n)C_n{x}^n\cdots \left(B\right)$
Multiplying (A) and (B)we get ${\left[(1+x)(1+x)\right]}^n=(C_0{x}^n+C_1{x}^{n-1}+\cdots +C_n{x}^0)$$\times (C_0-C_{1}x+C_2{x}^2+\cdots +{(-1)}^n{C}_n{x}^n)$
${(1+x^2)}^n=A_1{x}^0+A_2{x}^1+A_3{x}^2+\cdots x^n(C_0^2-C_1^2+\cdots +{(-1)}^n{C}_n^2)+\cdots +C_0{C}_n{(-1)}^n{x}^{2n}$
Coefficient of $x^n$ in R.H.S.$=C_0^2-C_1^2+C_2^2-C_3^2+\cdots {(-1)}^n{C}_n^2$

Now for the coefficient of $x^n$ in L.H.S
Let us consider the general term in L.H.S.
$T_{r+1}{=}^n{C}_r{(-x^2)}^r{=}^n{C}_r{x}^{2r}{(-1)}^r$ $\therefore$ for $x^n$ putting $2r=n\Rightarrow$ $r=\frac{n}{2}$

${\therefore }^T(\frac{n}{2}+1){=}^n{C}_{\frac{n}{2}}{(-1)}^{\frac{n}{2}}{x}^n$
$\therefore$ coefficient of $x^n$ in L.H.s.${=}^n{C}_{\frac{n}{2}}{(-1)}^{\frac{n}{2}}$ $=\{\begin{array}{cc}\frac{n!}{\frac{n}{2}!\frac{n}{2}!}& {(-1)}^{\frac{n}{2}}\text{if}n=\text{even}\\ 0& \text{if}n=\text{odd}\end{array}$ (as fractional factorial can not defined ) Now In our problem $n=15$ which is odd $\therefore C_0^2-C_1^2+C_2^2+\cdots +{(-1)}^{15}{C}_{15}^2=0$ or $\therefore C_0^2-C_1^2+C_2^2-C_3^2+\cdots -C_{15}^2=0$