Question

设S1=1+1/12+1/22,S2=1+1/22+1/32 ...Sn=1+1/n2+1/(n+1)2,S=√S1+√S2+…+√Sn ,那么S=

设S1=1+1/12+1/22，S2=1+1/22+1/32...Sn=1+1/n2+1/(n+1)2,S=√S1+√S2+…+√Sn,那么S=
谢谢

Answer 1

你可以先试试数学归纳法，直接法我先想想，晚上上图

追问

谢谢，我已经解出了，答案是n加1分之n平方+2n,对了提问数字后面的2是平方，打错了

追答

有一个小疏忽，最后答案后面再加个n

Answer 2

Sn=1+1/n^2+1/(n+1)^2=(n^4+2n^3+3n^2+2n+1)/(n^2*(n+1)^2)=(n*(n+1)+1)^2/(n^2*(n+1)^2)
故√Sn=√(n*(n+1)+1)^2/(n^2*(n+1)^2)=[n(n+1)+1]/[n(n+1)]=1+1/[n(n+1)]=1+1/n-1/(n+1)
所以:
√S1=1+1-1/2
√S2=1+1/2-1/3
√S3=1+1/3-1/4
....
√Sn=1+1/n-1/(n+1)
s=n+1-1/(n+1)