Question

设{an}为等差数列,Sn为其前n项和,已知S5=5,s11=77,Tn为数列｛Sn/n}的前n项和,求Tn

Answer 1

S5=5a1+10d=5 a1+2d=1

S11=11a1+55d=77 a1+5d=7

d=2 a1=-3

Sn=na1+n(n-1)d=-3n+n(n-1)=n^2-4n

令bn=Sn/n=n-4 {bn}是等差数列

b1=-3,d=1

Tn=nb1+n(n-1)d=-3n+n(n-1)/2=(n^2-7n)/2