the value of $\displaystyle \sum^{3m}_{r=0}(-1)^r\binom{6m}{2r}$ is
what i try
opening sum
$$\binom{6m}{0}-\binom{6m}{2}+\binom{6m}{4}-\binom{6m}{6}+\cdots +(-1)^{3m}\binom{6m}{6m}$$
$$(1+x)^{6m}=\binom{6m}{0}+\binom{6m}{1}x+\binom{6m}{2}x^2+\cdots +\binom{6m}{6m}x^{6m}$$
$$(1-x)^{6m}=\binom{6m}{0}-\binom{6m}{1}x+\binom{6m}{2}x^2-\cdots +(-1)^m\binom{6m}{6m}x^{6m}$$
$$(1+x)^{6m}+(1-x)^{6m} = 2\bigg(\binom{6m}{0}+\binom{6m}{2}x^2+\binom{6m}{4}x^4+\cdots\bigg)$$
how do i solve it help me please
Your value$=\frac{(1+i)^{6m}+(1-i)^{6m}}{2}$, here $i^2=-1$.