If $(1+x)^n=^n\!\!C_0+^n\!\!C_1x+^n\!\!C_2x^2+\cdots+^n\!\!C_nx^n$, then show that
$$(^n\!\!C_0-^n\!\!C_2+^n\!\!C_4-^n\!\!C_6+\cdots)^2+(^n\!C_1-^n\!\!C_3+^n\!\!C_5-\cdots)^2\\
=^n\!\!C_0+^n\!\!C_1+^n\!\!C_2+\cdots+^n\!\!C_n$$
We have by putting, $x=1$,
$2^n=^n\!\!C_0+^n\!\!C_1+^n\!\!C_2+\cdots+^n\!\!C_n$.
Putting , $x=-1$, we get $0=^n\!\!C_0-^n\!\!C_1+^n\!\!C_2-\cdots+(-1)^n \,^n\!\!C_n$. But how can I show that result? Thanks in advance.
