求和[x+(1/y)]+[x^2+(1/y^2)]+...+[x^n+(1/y^n)] (其中x≠0,x≠0,y≠1) 急!!

[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^2+……+x^n是以x为首项，x为公比的等比数列，有x项
所以和=x(x^n-1)/(x-1)

1/y+1/y^2+……+1/y^n是以1/y为首项，1/y为公比的等比数列，有x项
所以和=(1/y)[(1/y)^n-1]/(1/y-1)
=(1-y)[(1/y)^n-1]

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

=(x+xx+xxx+……+x^n)+(1/y+1/yy+……+1/y^n)
=x*(1-x^n)/(1-x)+1/y*(1-1/y^n)/(1-1/y)
=x*(1-x^n)/(1-x)+(1-1/y^n)/(y-1)