留名等等帮解

n^2<n(n+1)<(n+1)^2
n<√n^2+n<n+1
S1=9/4, S2=49/36, S3=169/144, ...,
Sk
=1+1/k^2+1/(k+1)^2
=(k^4+2k^3+3k^2+2k+1)/[k^2(k+1)^2]
=(k^2+k+1)^2/[k^2(k+1)^2]
√Sk=(k^2+k+1)/k(k+1)=1+1/k(k+1)=1+[1/k-1/(k+1)]
S=√S1+√S2+...+√Sn=n+1-1/(n+1)

我觉得是n。因为根据例题推演，假设一个新值为4，那么就有√4×4+4=√20，且4＜√20＜5所以√4×4+4整数部分为4，按顺序依此类推，则有，n＜√n×n+n＜n+1，所以n整数部分为n