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1+1/x+1/x^2+1/x^3+...1/x^n=[1-(1/x)^(n+1)]/[1-(1/x)]
题目详情
1+1/x+1/x^2+1/x^3+...1/x^n=[1-(1/x)^(n+1)]/[1-(1/x)]
▼优质解答
答案和解析
将左边的多项式乘以1/x,即
(1/x)(1+1/x+1/x^2+1/x^3+...1/x^n)
=1/x+1/x^2+1/x^3+...1/x^n+1/x^(n+1)
再将左边的多项式减去上面的结果,得
(1+1/x+1/x^2+1/x^3+...1/x^n)-(1/x)(1+1/x+1/x^2+1/x^3+...1/x^n)
=1+(1/x+1/x^2+1/x^3+...1/x^n)-(1/x+1/x^2+1/x^3+...1/x^n)-1/x^(n+1)
=1-1/x^(n+1)
即：
1+1/x+1/x^2+1/x^3+...1/x^n=[1-(1/x)^(n+1)]/[1-(1/x)]
(1/x)(1+1/x+1/x^2+1/x^3+...1/x^n)
=1/x+1/x^2+1/x^3+...1/x^n+1/x^(n+1)
再将左边的多项式减去上面的结果,得
(1+1/x+1/x^2+1/x^3+...1/x^n)-(1/x)(1+1/x+1/x^2+1/x^3+...1/x^n)
=1+(1/x+1/x^2+1/x^3+...1/x^n)-(1/x+1/x^2+1/x^3+...1/x^n)-1/x^(n+1)
=1-1/x^(n+1)
即：
1+1/x+1/x^2+1/x^3+...1/x^n=[1-(1/x)^(n+1)]/[1-(1/x)]
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