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)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？S3n=3na1+3n(3n-1)d/23(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