数列{An}的前n项和Sn=2n^2-3n,Bn=An*2^n,求{Bn}的前n项和Tn

来源：学生作业学帮网编辑：学帮网时间：2024/06/09 05:15:26

Sn=2n^2-3n
S(n-1)=2(n-1)^2-3(n-1)
想减得：An=4n-5
Bn=An*2^n=（4n-5）2^n
Tn=-2+3*2^2+……+(4n-5)2^n
2Tn= -2^2+……+(4n-9)2^n+(4n-5)2^(n+1)
想减得：Tn=（n+1)2^(n+3)-5*2^(n+1)-14
我的答案又对又快,求你了,亲.

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