问题
选择题
已知数列{an}的前n项和Sn=(n2+n) 2n,则数列{
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答案
∵Sn=(n2+n)﹒2n,
∴Sn-1=[(n-1)2+(n-1)]﹒2n-1,(n≥2)
两式相减可得,sn-sn-1=(n2+n)﹒2n-[(n-1)2+(n-1)]﹒2n-1,
=2n-1•(n2+3n)(n≥2)
n=1时,a1=s1=4适合上式
∴an=2n-1•(n2+3n)
∴
=(n+3)•2n-1an n
∴sn=4•20+5•2+…+(n+3)•2n-1
2sn=4•2+5•21+…+(n+2)•2n-1+(n+3)•2n
两式相减可得,-sn=4+2+22+…+2n-1-(n+3)•2n
=4+
-(n+3)•2n2(1-2n-1) 1-2
=4+2n-2-(n+3)•2n
=2-(n+2)•2n
∴Sn=(n+2)•2n-2
故选C.