Sn=z＋z^2＋z^3＋...+z^n,zSn=z^2＋z^3＋z^4+...+z^(n+1).两式相减,得(1-z)Sn=(z＋z^2＋z^3＋...+z^n)-[z^2＋z^3＋z^4+...+z^(n+1)]=z+(z^2-z^2)+(z^3-z^3)+...(z^n-z^n)-z^(n+1)=z-z^(n+1),故Sn=[z-z^(n+1)]/(1-z)=...