請求幫幫忙二項式C(64,0)+C(64,4)+...

(1+x)^n=C(n,0)+C(n,1)x+C(n,2)x^2+...
(1+1)^n=C(n,0)+C(n,1)*1+C(n,2)*1^2+..... ----(A)
(1+i)^n=C(n,0)+C(n,1)*i+C(n,2)*i^2+.... ----(B)
(1-1)^n=C(n,0)+C(n,1)*i^2+C(n,2)*i^4+.... ----(C)
(1-i)^n= C(n,0)+C(n,2)*i^3+C(n,2)*i^6+.... ----(D)
[(A)+(B)+(C)+(D)]/4 obtains
[2^n+(1+i)^n+(1-i)^n]/4= C(n,0)+C(n,4)+C(n,8)+....
[2^n+2^(1+ n/2) cos(nπ/4)]/4=C(n,0)+C(n,4)+C(n,8)+....
so, C(n,0)+C(n,4)+C(n,8)+...= 2^(n-2)+ 2^(n/2 - 1)cos(n π/4)

Check!
n=4: C(4,0)+C(4,4)=2
2^(4-2)+2^(4/2 -1)cos(π)=4-2=2, right!
n=8: C(8,0)+C(8,4)+C(8,8)=1+70+1=72
2^(8-2)+2^(8/2 -1)cos(2π)=64+8=72, right!
....

Note: C(n,k)*1^k+C(n,k)*i^k+C(n,k)*i^(2k)+C(n,k)*i^(3k)
=C(n,k)[1+i^k+(-1)^k+(-i)^k]
If k≡0 (mod4), then 1+i^k+(-1)^k+(-i)^k=4
If k≡1 (mod4), then 1+i^k+(-1)^k+(-i)^k=1+i-1-i=0
If k≡2 (mod4), then 1+i^k+(-1)^k+(-i)^k=1-1+1-1=0
If k≡3 (mod4), then 1+i^k+(-1)^k+(-i)^k=1-i-1+i=0

2010-05-17 20:48:45 補充：
In this problem,n=64
Answer=2^62+2^31= 2^31 (2^31 + 1)

2010-05-17 20:59:16 補充：
2^62+2^31=4611686020574871552

這個解法真是漂亮!!!!
如果要求諸如C(27,0)+C(27,3)+C(27,6)+....C(27,27)
應該可以利用ω的性質了