1

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

jacky
  • 5,194
  • 1
    the coefficients are 1,0,-1,0,1,0,-1,0,... of period 4, so can take x, such that $x^4=1$ – Bonbon Feb 04 '19 at 14:11
  • 1
    Use $(-1)^r=i^{2r}$ and the sum is half terms of the binomial expansion ... well, the answer is below already. – rtybase Feb 04 '19 at 14:12

1 Answers1

3

Your value$=\frac{(1+i)^{6m}+(1-i)^{6m}}{2}$, here $i^2=-1$.

Bonbon
  • 891