Question

已知数列{an}的通项公式为an=(1-2n)*(-1/2)^(n+1),求前n项和

Answer 1

Sn-S(n-1)=(1-2n)*(-1/2)^(n+1)
S2-S1=(-3)*(-1/2)^3
S3-S2=(-5)*(-1/2)^4
S4-S3=(-7)*(-1/2)^5
Sn-S(n-1)=(1-2n)*(-1/2)^(n+1)
全部相加
S1-Sn=3*(-1/2)^3+5*(-1/2)^4+7*(-1/2)^5+……+(2n-1)*(-1/2)^(n+1)
(-1/2)(S1-Sn)=3*(-1/2)^4+5*(-1/2)^5+7*(-1/2)^6+……+(2n-1)*(-1/2)^(n+2)
两式相减
(3/2)(S1-Sn)=3*(-1/2)^3+2[(-1/2)^4+(-1/2)^5+(-1/2)^6+……+(-1/2)^(n+1)]-(2n-1)*(-1/2)^(n+2)
S1-Sn=2*(-1/2)^3+[1/4-(-1/2)^n]/9-(2/3)(2n-1)*(-1/2)^(n+2)
Sn=S1-2*(-1/2)^3-[1/4-(-1/2)^n]/9+(2/3)(2n-1)*(-1/2)^(n+2)
Sn=-1/4-2*(-1/2)^3-[1/4-(-1/2)^n]/9+(2/3)(2n-1)*(-1/2)^(n+2)

佩服自己，好有耐心