Question

已知数列{an}是an=2^n,bn=3n+1的等比数列,Cn=(3n+1)*2^n求Sn.要完整过程.

Answer 1

{an}是等比数列,｛bn}是等差数列

｛cn}是“差比积”数列,求和方法是错位相减

Sn=4×2+7×2²+10×2³+.+(3n+1)2^n ①

两边同时乘以2:

2Sn=4×2²+7×2³+10×2⁴+.+(3n-2)2^n+(3n+1)*2^(n+1） ②

①-②：

-Sn=8+3×2²+3×2³+.+3×2^n-(3n+1)*2^(n+1)

=8+3(2²+2³+.+2^n)-(3n+1)*2^(n+1)

=8+3×4[2^(n-1)-1]/(2-1)-(3n+1)*2^(n+1)

=-4+3*2^(n+1)-(3n+1)*2^(n+1)

=-4-(3n-2)*2^(n+1)

∴Sn=(3n-2)*2^(n+1)+4