Find the value of $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$
I found the general term of the sequence.
It is $\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}$
So the sequence becomes $\sum_{k=1}^{1999}\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}$
I tried telescoping but i could not split it into two partial fractions.And this raised to $\frac{1}{2}$ is also troubling me.What should i do to find the answer?
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Vinod Kumar Punia
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For general term express it in form $p/q$ – Archis Welankar Dec 30 '15 at 07:32
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2Note: $1 + 1/k^2 + 1/(k+1)^2 = \frac {(k(k+1) + 1)^2}{(k(k+1))^2}$ – fleablood Dec 30 '15 at 07:36
2 Answers
13
After taking LCM, we get the general term of the series as:
$$\sqrt{\frac{(k^{2}+k+1)^{2}}{k^{2}(k+1)^{2}}} $$
$$=> \frac{k^{2}+k+1}{k^{2}+k}$$
$$=> 1 + \frac{1}{k^{2}+k}$$
So we have $$\sum_{k=1}^{1999} 1 + \frac{1}{k^{2}+k}$$
$$=> 1999 + \sum_{k=1}^{1999}\frac{1}{k(k+1)}$$
$$=> 1999 + \sum_{k=1}^{1999}\frac{1}{k} - \frac{1}{(k+1)}$$
$$=> 1999 + 1 - \frac{1}{2000}$$
$$=> 2000 - \frac{1}{2000}$$
Ashish Gupta
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Given $$1+\frac{1}{k^2}+\frac{1}{(k+1)^2} = 1+\frac{1}{k^2}+\frac{1}{(k+1)^2}-\frac{2}{k(k+1)}+\frac{2}{k(k+1)}$$
So $$1^2+\left[\frac{1}{k}-\frac{1}{(k+1)}\right]^2+2\left[\frac{1}{k}-\frac{1}{(k+1)}\right]=\left[1+\frac{1}{k}-\frac{1}{k+1}\right]^2$$
juantheron
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