TITLE: A pencil approach to find $\sum \limits_{i=1}^{69} \sqrt{\left( 1+\frac{1}{i^2}+\frac{1}{(i+1)^2}\right)}$
QUESTION [36 upvotes]: What is the fastest, paper-pencil method of finding $$\sum \limits_{i=1}^{69} \sqrt{\left( 1+\frac{1}{i^2}+\frac{1}{(i+1)^2}\right)}?$$
This is actually a quantitative aptitude problem, and hence the solutions should be fast enough and probably under a minute.
Using wolframalpha, the sum seems to be $\frac{4899}{70}$; however, I am not sure how to find this quickly in the paper-pencil way.

REPLY [4 votes]: This is a late response (since I joined three weeks ago) but here is another method. Write
\begin{align}
1 + \frac{1}{i^2} + \frac{1}{(i + 1)^2} &= 1 + \frac{2}{i(i+1)} + \left(\frac{1}{i^2} - \frac{2}{i(i+1)} + \frac{1}{(i+1)^2}\right)\\
&= 1 + 2\left(\frac{1}{i} - \frac{1}{i+1}\right) + \left(\frac{1}{i} - \frac{1}{i+1}\right)^2\\
&= \left(1 + \frac{1}{i} - \frac{1}{i+1}\right)^2.
\end{align}
Then
\begin{equation}
\sum_{i = 1}^{69} \sqrt{1 + \frac{1}{i} + \frac{1}{i^2}} = \sum_{i = 1}^{69} \left(1 + \frac{1}{i} - \frac{1}{i + 1}\right) = 69 + \sum_{i = 1}^{69} \left(\frac{1}{i} - \frac{1}{i+1}\right) = 70 - \frac{1}{70} = \frac{4899}{70}.
\end{equation}