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Sum of Square Root Series

CAT 2008 Slot 1 · Quantitative Ability · Hard · Algebra

This is a hard Quantitative Ability question from the CAT 2008 Slot 1 paper. It tests Algebra. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Find the sum 1+112+122+1+122+132+⋯+1+120072+120082\sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}} + \sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}} + \dots + \sqrt{1 + \frac{1}{2007^2} + \frac{1}{2008^2}}

  1. A.

    2008−120082008 - \frac{1}{2008}

  2. B.

    2007−120072007 - \frac{1}{2007}

  3. C.

    2007−120082007 - \frac{1}{2008}

  4. D.

    2008−120072008 - \frac{1}{2007}

  5. E.

    2008−120092008 - \frac{1}{2009}

Answer

A

Explanation

General term Tn=1+1n2+1(n+1)2T_n = \sqrt{1 + \frac{1}{n^2} + \frac{1}{(n+1)^2}}.

Simplifying inside the square root: 1+1n2+1(n+1)2=n2(n+1)2+(n+1)2+n2n2(n+1)2=n4+2n3+3n2+2n+1n2(n+1)2=(n2+n+1)2n2(n+1)21 + \frac{1}{n^2} + \frac{1}{(n+1)^2} = \frac{n^2(n+1)^2 + (n+1)^2 + n^2}{n^2(n+1)^2} = \frac{n^4 + 2n^3 + 3n^2 + 2n + 1}{n^2(n+1)^2} = \frac{(n^2 + n + 1)^2}{n^2(n+1)^2}

So Tn=n2+n+1n(n+1)=1+1n(n+1)=1+1n−1n+1T_n = \frac{n^2 + n + 1}{n(n+1)} = 1 + \frac{1}{n(n+1)} = 1 + \frac{1}{n} - \frac{1}{n+1}.

Sum S=∑n=12007Tn=∑n=12007(1+1n−1n+1)=2007+(1−12008)=2008−12008S = \sum_{n=1}^{2007} T_n = \sum_{n=1}^{2007} \left(1 + \frac{1}{n} - \frac{1}{n+1}\right) = 2007 + \left(1 - \frac{1}{2008}\right) = 2008 - \frac{1}{2008}.

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