Question

数学问题，不懂，求高手解答，例子如下：

Answer 1

√ [1+1/n^2 + 1/(n+1)^2 ]
=√[1+(n^2+n^2+2n+1)/n^2(n+1)^2]
=√ [1+2(n^2+n)/n^2(n+1)^2 + 1/n^2(n+1)^2]
=√[1+2/n(n+1) + 1/n^2(n+1)^2]
=√{[1+1/n(n+1)]^2}
=1+1/n(n+1)
=1+1/n-1/(n+1)
所以原式=1+1/1-1/2+1+1/2-1/3...+1+1/99-1/100
=99*1+1-1/100=9999/100=99.99

Answer 2

1+1/i^2+1/(i+1)^2={(i+1)^2+i^2+[i^2*(i+1)^2]}/[i^2*(i+1)^2]
=(i^4+2i^3+3i^2+2i+1)/[i^2*(i+1)^2]
=(i^2+i+1)^2/(i^2)/(i+1)^2
从而每一项根号之后都是
(i^2+i+1)/i(i+1)=1+1/i(i+1)=1+1/i-1/(i+1)
所以总共99项，答案是
99+1/1-1/100=9999/100