已知数列{an}的通项公式为an=(-1)^n+1(3n-2),求前100项的和

a1=3-2
a2=-(6-2)=-3*2+2
a3=9-2=3*3-2
a4=-(12-2)=-3*4+2
.
a99=3*99-2
a100=-3*100+2
S100=a1+a2+a3+...+a99+a100
=3-2-3*2+2+3*3-2-3*4+2+.+3*99-2-3*100+2
=3-3*2+3*3-3*4+3*5-3*6+...+3*99-3*100
=-3-3-3-3-3-3...-3共50个
=-150

an=(-1)^n+1(3n-2)
sn=-1(1-(-1)^n)/(1-(-1)) + n(3+3n)/2-2n
=-1/2+(-1)^n+(3n^2-n)/2
s100=-0.5+1+14950=14950.5