求(r^4+r^2+1)/(r^4+r)的前n项和

Question

求(r^4+r^2+1)/(r^4+r)的前n项和

Answer 1

就都用n来写吧
由公式可知
n^4+n^2+1=(n^6-1)/(n^2-1)
=(n^3+1)(n^3-1)/(n^2-1)
=(n^3+1)(n^2+n+1)/(n+1)
n^4+n=n(n^3+1)
故(n^4+n^2+1)/(n^4+n)=(n^2+n+1)/(n^2+n)
=1+1/n(n+1)
=1+ 1/n - 1/(n+1)
所以其前n项和Sn=1+1/1-1/2 + 1+1/2-1/3 +……+ 1+1/n - 1/(n+1)
=n+1- 1/(n+1)