A pencil approach to find $\sum \limits_{i=1}^{69} \sqrt{\left( 1+\frac{1}{i^2}+\frac{1}{(i+1)^2}\right)}$ 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.
 A: Edit: Now in beautiful high-definition technicolor!
$$\sum \limits_{k=1}^n \sqrt{\color{red}1+\color{Green}{\frac{1}{k^2}}+\color{Blue}{\frac{1}{(k+1)^2}}}$$
$$=\sum_{k=1}^n\sqrt{\frac{\color{red}{k^2(k+1)^2}+\color{Green}{(k+1)^2}+\color{Blue}{k^2}}{k^2(k+1)^2}}$$
$$=\sum_{l=1}^n \sqrt{\frac{k^4+2k^3+3k^2+2k+1}{k^2(k+1)^2}}$$
$$=\sum_{k=1}^n\sqrt{\frac{\color{Red}1k^4+(\color{Red}1+\color{Green}1)k^3+(\color{Red}1+\color{Green}1+\color{Blue}1)k^2+(\color{Green}1+\color{Blue}1)k+\color{Blue}1}{k^2(k+1)^2}} $$
$$=\sum_{k=1}^n\sqrt{\frac{\color{Red}{k^2(k^2+k+1)}+\color{Green}{k(k^2+k+1)}+\color{Blue}{(k^2+k+1)}}{k^2(k+1)^2}} $$
$$=\sum_{k=1}^n\sqrt{\frac{(k^2+k+1)^2}{k^2(k+1)^2}}$$
$$=\sum_{k=1}^n\frac{\color{Purple}{k^2+k}+\color{Blue}1}{k^2+k}$$
$$=\sum_{k=1}^n \left(\color{Purple}1+\color{Blue}{\frac{1}{k(k+1)}}\right) $$
$$=\sum_{k=1}^n \left(1+\frac{1}{k}-\frac{1}{k+1}\right)$$ 
$$=n+1-\frac{1}{n+1}.$$
Specializing $n=69$ gives $69+1-1/70=4899/70$. Note that it's highly useful to memorize this:
$$\frac{1}{k(k+a)}=\frac{1}{a}\frac{(k+a)-k}{k(k+a)}=\frac{1}{a}\left(\frac{1}{k}-\frac{1}{k+a}\right).$$
A: 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}
A: By multiplying to get a common denominator we have that $$1+\frac{1}{i^{2}}+\frac{1}{(i+1)^{2}}=\frac{i^{2}(i+1)^{2}+(i+1)^{2}+i^{2}}{i^{2}(i+1)^{2}} $$
$$=\frac{i^{4}+2i^{3}+3i^{2}+2i+1}{i^{2}(i+1)^{2}}=\frac{(i^{2}+i+1)^{2}}{i^{2}(i+1)^{2}}.$$ Hence our sum is $$\sum_{i=1}^{69}\frac{i^{2}+i+1}{i(i+1)}=\sum_{i=1}^{69}\left(\frac{i^{2}+2i+1}{i(i+1)}-\frac{i}{i(i+1)}\right) $$
$$=\sum_{i=1}^{69}\frac{i+1}{i}-\frac{1}{i+1}=69+\sum_{i=1}^{69}\left(\frac{1}{i}-\frac{1}{i+1}\right).$$  The right hand side telescopes, so we conclude that the answer is $$69+1-\frac{1}{70}=70-\frac{1}{70}=\frac{4899}{70}.$$
