Question

已知{an}是等差数列，前n项和记为Sn，已知数列Sm，求证：Sm，S2m-Sm，S3m-S2m也成等差数列

Answer 1

S2m-Sm
=(a1+a2+……+a2m)-(a1+a2+……+am)
=a(m+1)+a(m+2)+……+a2m

同理
S3m-S2m
=a(2m+1)+a(2m+2)+……+a3m

所以(S2m-Sm)-Sm
=[a(m+1)+a(m+2)+……+a2m]-(a1+a2+……+am)
=[a(m+1)-a1]+[a(m+2)-a2]+……+(a2m-am)
=md+md+……+md
=m²d

(S3m-S2m)-(S2m-Sm)
=[a(2m+1)+a(2m+2)+……+a3m]-[a(m+1)+a(m+2)+……+a2m]
=[a(2m+1)-a(m+1)]+[a(2m+2)-a(m+2)]+……+(a3m-a2m)
=md+md+……+md
=m²d

所以(S2m-Sm)-Sm=(S3m-S2m)-(S2m-Sm)
所以Sm，S2m-Sm，S3m-S2m成等差数列

Answer 2

Sm=a1+....+am
S2m-Sm=a(m+1)+.....+a(2m)=a1+....+am+m^2*d
S3m-S2m=a(2m+1)+.....+a(3m)=a(m+1)+.....+a(2m)+m^2*d
所以等差