已知数列an的通项为an=(-1)^n*(2n-1)*cos(nπ/2)+1,前n项和为Sn,求S50=

题目

已知数列an的通项为an=(-1)^n*(2n-1)*cos(nπ/2)+1,前n项和为Sn,求S50=

答案

很古老的典型题目了.
令k∈N
n=4k+1时,cos(nπ/2)=0
n=4k+2时,cos(nπ/2)=-1
n=4k+3时,cos(nπ/2)=0
n=4k+4时,cos(nπ/2)=1
即从第1项开始,cos(nπ/2)按0,-1,0,1循环,每4项循环一次.奇数项为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,循环12次余两项,k从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