求和：（x+1/y)+[x^2+1/(y^2)]...+[x^n+1/(y^n)](x,y不等于1）我知道公式的但是不晓得这么化简、要详略步骤
2019-05-27
求和：（x+1/y)+[x^2+1/(y^2)]...+[x^n+1/(y^n)](x,y不等于1）
我知道公式的但是不晓得这么化简、要详略步骤
优质解答
(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)/(x-1)+(1/y)(1-1/y^n)/(1-1/y)
=x(x^n-1)/(x-1)+(1/y)[(y^n-1)/y^n]/[(y-1)/y]
=x(x^n-1)/(x-1)+(1/y)[(y^n-1)/y^n]*[y/(y-1)]
=x(x^n-1)/(x-1)+(y^n-1)/[(y^n)(y-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^n-1)/(x-1)+(1/y)(1-1/y^n)/(1-1/y)
=x(x^n-1)/(x-1)+(1/y)[(y^n-1)/y^n]/[(y-1)/y]
=x(x^n-1)/(x-1)+(1/y)[(y^n-1)/y^n]*[y/(y-1)]
=x(x^n-1)/(x-1)+(y^n-1)/[(y^n)(y-1)]