An algebra problem by Tarit Goswami

Algebra Level 4

$\small \sqrt{1+\frac 1{1^2} +\frac 1{2^2}} + \sqrt{1+\frac 1{2^2} +\frac 1{3^2}} + \sqrt{1+\frac 1{3^2} + \frac 1{4^2}}+\cdots + \sqrt{1+ \frac 1{1999^2} + \frac 1{2000^2}}= \, ?$

The answer is 1999.9995.

1 solution

Chew-Seong Cheong
Aug 15, 2016

$\begin{aligned} S & = \sum_{n=1}^{1999} \sqrt{1+\frac 1{n^2} + \frac 1{(n+1)^2}} \\ & = \sum_{n=1}^{1999} \sqrt{\frac{n^2(n+1)^2+(n+1)^2+n^2}{n^2(n+1)^2}} \\ & = \sum_{n=1}^{1999} \sqrt{\frac{n^2(n^2+2n+1)^2+n^2+2n+1+n^2}{n^2(n+1)^2}} \\ & = \sum_{n=1}^{1999} \sqrt{\frac{n^4+2n^3+3n^2+2n+1}{n^2(n+1)^2}} \\ & = \sum_{n=1}^{1999} \sqrt{\frac{(n^2+n+1)^2}{n^2(n+1)^2}} \\ & = \sum_{n=1}^{1999} \frac{n^2+n+1}{n(n+1)} \\ & = \sum_{n=1}^{1999} \frac{n^2+n+1}{n^2+n} \\ & = \sum_{n=1}^{1999} \left(1 + \frac 1{n(n+1)} \right) \\ & = \sum_{n=1}^{1999} \left(1 + \frac 1n - \frac 1{n+1} \right) \\ & = 1999 + \frac 11 - \frac 1{2000} \\ & = 1999 + \frac {1999}{2000} \\ & = \boxed{1999.9995} \end{aligned}$

Your solution is good... but if we see on the pattern of the n th term then we can see sum of n th terms will be (n^2-1)/n

Tarit Goswami - 4 years, 10 months ago

An algebra problem by Tarit Goswami

The answer is 1999.9995.

1 solution

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