Question

If $c_0,c_1,c_2,.......c_n$ denote the coefficients in the expansion of ${(1+x)}^n$, prove that
${c_0}^2+{c_1}^2+{c_2}^2+....{c_n}^2=\frac{|\underset{–}{2n}}{|\underset{–}{n}|\underset{–}{n}}$.

Answer 1

$⟹{(1+x)}^n=C_0+C_{1}x+C_{2}x+.......+C_n\cdot x^n⟶\left(I\right)$
${(x+1)}^n=C_0{x}^n+C_1{x}^{n-1}+.......C_n{x}^0⟶\left(II\right)$
Multiplying $\left(I\right)$ and $\left(II\right)$
$C_0^2+C_1^2+C_2^2+......+C_{n-1}^2+C_n^2$
$=$ coefficient of $x^n$ in ${(1+x)}^{2n}$
${=}^{2n}{C}_n$
$=\frac{\left(2n\right)!}{n!n!}$