已知数列an=n,前n项和为Sn,bn=n³,前n项和为Tn,求证Tn=Sn²

题目

已知数列an=n,前n项和为Sn,bn=n³,前n项和为Tn,求证Tn=Sn²

答案

可以用归纳法
T1=S1`2
设n=k 时成立
n=k+1
T（k+1）=Tk+（k+1）`3=Sk`2+(k+1)`3
S(k+1)`2=(Sk+k+1)`2
=Sk`2+2(k+1)Sk+(k+1)`2
=Sk`2+2(K+1)(k+1)k/2+(k+1)`2 注：Sn=（n+1）n/2
=Sk`2+k(k+1)`2+(k+1)`2
=Sk`2+(k+1)`3
=T（k+1）