Question

已知an是等差数列,前n项和为Sn,求证:S3n=3(S2n-Sn)

Answer 1

证明：设等差数列{an}公差为d，由定义知
Sn=a1+a2+...+an
S2n=a1+a2+...+an+a(n+1)+...+a2n=Sn+a1+nd+a2+nd+...+an+nd=Sn+Sn+n*nd=2Sn+n*nd，
可得n*nd=S2n-2Sn，所以
S3n=a1+a2+...+a2n+a(2n+1)+...+a3n
=S2n+a1+2nd+a2+2nd+...+an+2nd=S2n+Sn+n*2nd=S2n+Sn+2*(n*nd)
=S2n+Sn+2(S2n-2Sn)=3(S2n-Sn).证毕。

Answer 2

S3n=3na1+3n(3n-1)d
3(S2n-Sn)=3(2na1+2n(2n-1)d/2-na1-n(n-1)d/2)
=3na1+3n(3n-1)d
所以S3n=3(S2n-Sn)

追问

3(S2n-Sn)=3(2na1+2n(2n-1)d/2-na1-n(n-1)d/2)
请问这一步怎么来的？为什么用除？

追答

S3n=3na1+3n(3n-1)d/2

3(S2n-Sn)=3(2na1+2n(2n-1)d/2-na1-n(n-1)d/2)
=3(2na1+(4n^2-2n)d/2-na1-(n^2-n)d/2)
=3(na1+(3n^2-n)d/2)
=3na1+3n(3n-1)d/2
所以S3n=3(S2n-Sn)

3(S2n-Sn)=3(2na1+2n(2n-1)d/2-na1-n(n-1)d/2)
这里分别用求和公式求出
S3n S2n Sn 得到