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Sn=z+z^2+z^3+…+z^n,zSn=z^2+z^3+z^4+…+z^(n+1).兩式相减,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)=…
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