问题补充:
填空题已知数列{an}的前n项和为Sn,Sn=2an-1,n∈N*,数列{(n+1)an}的前n项和为________.
答案:
n×2n解析分析:先确定数列的通项,再利用错位相减法求数列的和,即可得到结论.解答:由Sn=2an-1,得Sn-1=2an-1-1(n≥2).两式相减,得an=2an-2an-1,即an=2an-1(n≥2).∴数列{an}是公比为2的等比数列.又S1=2a1-1,所以a1=1.∴an=2n-1.∴数列{(n+1)an}的前n项和为Sn=2×1+3×2+4×22+…+(n+1)×2n-1,∴2Sn=2×2+3×22+4×23+…+(n+1)×2n,∴-Sn=2×1+2+22+…+2n-1-(n+1)×2n,∴Sn=n×2n,故