Question

等差数列an的公差为d前n项和为sn则数列sns 2 n sns 3ns2n也是公差d为

已知等差数列｛an｝的前n项和为Sn,公差为d.
求证：Sn,S2n-Sn,S3n-S2n,.成等差数列

Answer 1

｛an｝为察含悔等差数列,首项为a1,则
Sn=na1+n(n-1)d/2
S2n=2na1+2n(2n-1)d/2
因为,S2n-Sn=na1+n(3n-1)d/败正2
所以,(S2n-Sn)-Sn=n²d
因为,k>1时
[Skn -S(k-1)n]-[S(k-1)n -S(k-2)n]
={a[(k-1)n+1] +a[(k-1)n+2]+...+a[kn] } - {a[(k-2)n+1] +a[(k-2)n+2]+...+a[(k-1)n] }
={a[(k-1)n+1] -a[(k-2)n+1] }+ {a[(k-1)n+2] -a[(k-2)n+2]}+...+{a[kn] -a[(k-1)n] }
=nd+nd+...+nd （总共n项）
=n²d（为常数）
所以Sn、S2n-Sn、S3n-S2n、.成等差老简数列,公差为n²d