Question

1+1/x+1/x^2+1/x^3+...1/x^n=[1-(1/x)^(n+1)]/[1-(1/x)]

Answer 1

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)]