很古老的典型題目了.
令k∈N
n=4k+1時,cos(nπ/2)=0
n=4k+2時,cos(nπ/2)=-1
n=4k+3時,cos(nπ/2)=0
n=4k+4時,cos(nπ/2)=1
即從第1項開始,cos(nπ/2)按0,-1,0,1循環(huán),每4項循環(huán)一次.奇數(shù)項為0.
a(4k+1)+a(4k+2)+a(4k+3)+a(4k+4)
=(-1)^(4k+1)· [2(4k+1)-1]·0 +1+(-1)^(4k+2)· [2(4k+2)-1]·(-1)+1
+(-1)^(4k+3)· [2(4k+3)-1]·0 +1+(-1)^(4k+4)· [2(4k+4)-1]·1 +1
=16k+14
50÷4=12余2,循環(huán)12次余兩項,k從0到11
S50=16×(0+1+...+11)+14×12 +(-1)^49· (2×49-1)·cos(49π/2) +1+(-1)^50· (2×50-1)·cos(50π/2)+1
=16×11×12/2 +14×12 +1+99×(-1)+1
=1127