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求Sn=(x+1/y)+x^2+1/y^2)+·······（x^n+1/y^n)
题目详情
求Sn=(x+1/y)+x^2+1/y^2)+·······（x^n+1/y^n)
▼优质解答
答案和解析
Sn=(x+1/y)+x^2+1/y^2)+·······（x^n+1/y^n)
=(x+x^2+······+x^n)+(1/y+1/Y^2+······+1/y^n)
=[x-x^(n+1)]/(1-x)+[1/y-1/y^(n+1)]/(1-1/y)
=[x-x^(n+1)]/(1-x)+(y^n-1)/[y^(n+1)-y^n]
=(x+x^2+······+x^n)+(1/y+1/Y^2+······+1/y^n)
=[x-x^(n+1)]/(1-x)+[1/y-1/y^(n+1)]/(1-1/y)
=[x-x^(n+1)]/(1-x)+(y^n-1)/[y^(n+1)-y^n]
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