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當x=1時,sn=1+2+3+…+n=n(n+1)2;當x≠0且x≠1時,Sn=x+2x2+3x3+…+nxn,①xSn=x2+2x3+3x4+…+nxn+1,②①-②,得(1-x)Sn=x+x2+x3+…+xn-nxn+1,所以,sn=x(1−xn)(1−x)2-nxn+11−x.
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