已知an是等差数列,前n项和为Sn,求证:S3n=3(S2n-Sn)

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已知an是等差数列,前n项和为Sn,求证:S3n=3(S2n-Sn)

已知an是等差数列,前n项和为Sn,求证:S3n=3(S2n-Sn)
已知an是等差数列,前n项和为Sn,求证:S3n=3(S2n-Sn)

已知an是等差数列,前n项和为Sn,求证:S3n=3(S2n-Sn)
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)

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+...

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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。

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