Question

数列问题!bn=2^n*(2n+1),求bn的前n项和,这个用错位相减还是裂项相消?方法啊,求指导

Answer 1

错位相减

Sn=3*2^1+5*2^2+7*2^3……+(2n+1)*2^n

2Sn= 3*2^2+5*2^3+……+(2n-1)*2^n+(2n+1)*2^(n+1)

相减得

-Sn=6+2*2^2+2*2^3+……+2*2^n-(2n+1)*2^(n+1)

-Sn=6+2^(n+2)-8-(2n+1)*2^(n+1)=-(2n-1)*2^(n+1)-2

所以

Sn=(2n-1)*2^(n+1)-2