Question

已知数列通项公式，求数列前n 项和

Answer 1

Sn
(1/3)Sn
a1=2*(1/3)
(
1/
3a1)=2*(1/3)^2
a2=5*(1/3)^2
(1/3)
a2=5*(1/3)^3
a3=8*(1/3)^3
(1/3)
a3=8*(1/3)^4
------
--------
--------------
an-1=(3n-4)*(1/3)^n-1
(1/3)
an-1=(3n-4)*(1/3)^n
an=(3n-1)*(1/3)^n
(1/3)
an=(3n-1)*(1/3)^(n+1)
Sn-(1/3)Sn=2*(1/3)+(1/3)+(1/3)^2+---+(1/3)^(n-1)-(3n-1)*(1/3)^(n+1)
(左右错位相减)
(2/3)Sn=2/3+（1/3）*[1-(1/3)^(n-1)]/(1-1/3)-(3n-1)*(1/3)^(n+1)
(2/3)Sn=7/6-(1/2)(1/3)^(n-1)-(3n-1)(1/3)^(n+1)
Sn=7/4-(3/4)(1/3)^(n-1)-(3/2)(3n-1)(1/3)^(n+1)
方法正确计算可能有误
2与1方法相同把前边Sn乘以2再错位相减，得到
等比数列
自已
算吧