Question

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

Answer 1

[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]