Question

求数列2²+1/2²-1，3²+1/3²-1，...的前n项和

Answer 1

解答：
看通项an=[(n+1)²+1]/[(n+1)²-1]
=1+2/[(n+1)²-1]
=1+2/[(n+1-1)*(n+1+1)]
=1+2/[n(n+2)]
=1+1/n-1/(n+2)
∴ 前n项和=(1+1/1-1/3)+(1+1/2-1/4)+(1-1/3-1/5)+.......+[1+1/n-1/(n+2)]
=n+1-1/3+1/2-1/4+1/3-1/5+......+1/(n-1)-1/(n+1)+1/n-1/(n+2)
=n+1+1/2-1/(n+1)-1/(n+2)

Answer 2

an=[(n+1)²+1]/[(n+1)²-1]=1+2/[(n+1)²-1]=1+2/[n(n+2)]=1+[1/n-1/(n+2)]
a1=1+(1-1/3)
a2=1+(1/2-1/4)
a3=1+(1/3-1/5)
a4=1+(1/4-1/6)
......
an-1=1+[1/(n-1)-1/(n+1)]
an=1+[1/n-1/(n+2)]
Sn=n+1+1/2-1/(n+1)-1/n
=n+3/2-(2n+1)/[n(n+1)]