已知数列an的通项an=2的n次方cos(nπ),则 a1+a2+a3+...+a99+a100等于

cos[(2k-1)π]=cos(-π)=-1
cos(2kπ)=cos(2π)=1
即n为奇数时,cos(nπ)=-1,n为偶数时,cos(nπ)=1
a1+a2+a3+a4+...+a98+a99+a100
=-2+2²-2³+2⁴+...+2^98-2^99+2^100
=(2²+2⁴+...+2^100)-(2+2³+...+2^99) /看到了吧,拆成了两个等比数列求和
=4×(4^50 -1)/(4-1)-2×(4^50 -1)/(4-1)
=(4/3)×4^50 -4/3 -(2/3)×4^50 +2/3
=(2/3)×4^50 -2/3
=(2^101 -2)/3