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x=1の場合、sn=1+2+3+...+n=n(n+1)2;x=0の場合、x=1の場合、Sn=x+2x2+3x3+...+nxn,1xSn=x2+2x3x4+...+nxn+1,21-2,得(1-x)Sn=x+x2+x3+...+xn-nxn+1,だから、sn=x(1−xn)(1−x)2-nxn+11−x.
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