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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.

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are you missing the square root in the question? –  user13838 Oct 21 '11 at 19:11
    
@ percusse:Yes,Fixed now. –  Quixotic Oct 21 '11 at 19:11
    
your question is missing the square root edit: fixed –  opt Oct 21 '11 at 19:11
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@FoolForMath: It's not the first time I see you misspelling WolfRAmalpha. It has nothing to do with a wolf's arm (whatever that is) :) –  t.b. Oct 21 '11 at 19:17
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@t.b. I frequently see "intergral" on math forums... (and "symbols" in drumset classifieds, but that's a different story) –  The Chaz 2.0 Oct 21 '11 at 19:31

3 Answers 3

up vote 88 down vote accepted

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: $$\frac{1}{k(k+a)}=\frac{1}{a}\frac{(k+a)-k}{k(k+a)}=\frac{1}{a}\left(\frac{1}{k}-\frac{1}{k+a}\right).$$

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1  
I could not follow your answer from the first step itself.Could you please add the intermediate steps?Thanks. –  Quixotic Oct 21 '11 at 19:21
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@FoolForMath: I've added a number of intermediate steps. –  anon Oct 21 '11 at 19:35
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Me and anon's answers are really the same. I think we composed them at about the same time too. –  Eric Naslund Oct 21 '11 at 19:39
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That's beautiful,I never realized that latex could be so colorful.Aha,wish I could give you another +1 for "high-definition"! –  Quixotic Oct 21 '11 at 20:01
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$\color{Blue}{+}\color{Red}{1}$. :) –  Srivatsan Oct 22 '11 at 0:29

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 $$69+1-\frac{1}{70}=70-\frac{1}{70}=\frac{4899}{70}.$$

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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}

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