National Camp Exam 2018 P3

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Swapnil Barua
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National Camp Exam 2018 P3

Unread post by Swapnil Barua » Tue Apr 20, 2021 9:27 pm

Find x=$\sqrt[]{1+1/1²+1/2²}$ +$\sqrt[]{1+1/2²+1/3²}$...+$\sqrt[]{1+1/2012²+1/2013²}$

~Aurn0b~
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Re: National Camp Exam 2018 P3

Unread post by ~Aurn0b~ » Wed Apr 21, 2021 12:16 am

Swapnil Barua wrote:
Tue Apr 20, 2021 9:27 pm
Find x=$\sqrt[]{1+1/1²+1/2²}$ +$\sqrt[]{1+1/2²+1/3²}$...+$\sqrt[]{1+1/2012²+1/2013²}$
$\textbf{Solution}$
$$\sum_{n=1}^{2012}\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}$$

$$=\sum_{n=1}^{2012}\frac{\sqrt{n^2(n+1)^2+n^2+(n+1)^2}}{n(n+1)}$$

$$=\sum_{n=1}^{2012}\frac{\sqrt{n^2(n+1)^2+2n^2+2n+1}}{n(n+1)}$$

$$=\sum_{n=1}^{2012}\frac{\sqrt{n^2(n+1)^2+2n(n+1)+1}}{n(n+1)}$$

$$=\sum_{n=1}^{2012}\frac{\sqrt{(n(n+1)+1)^2}}{n(n+1)}$$

$$=\sum_{n=1}^{2012}\frac{n(n+1)+1}{n(n+1)}$$

$$=\sum_{n=1}^{2012} 1+\sum_{n=1}^{2012}\frac{1}{n(n+1)}$$

$$=\sum_{n=1}^{2012} 1+\sum_{n=1}^{2012}(\frac{1}{n}-\frac{1}{n+1})$$

$$=2012+(1-\frac{1}{2013})=\frac{4052168}{2013}.\blacksquare$$

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Swapnil Barua
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Re: National Camp Exam 2018 P3

Unread post by Swapnil Barua » Wed Apr 21, 2021 8:16 am

You did the summation of infinite series?

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Mehrab4226
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Re: National Camp Exam 2018 P3

Unread post by Mehrab4226 » Wed Apr 21, 2021 12:31 pm

Swapnil Barua wrote:
Wed Apr 21, 2021 8:16 am
You did the summation of infinite series?
No, he did the sum of $\frac{1}{n}-\frac{1}{n+1}$. He probably used Telescoping to find that.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

~Aurn0b~
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Re: National Camp Exam 2018 P3

Unread post by ~Aurn0b~ » Wed Apr 21, 2021 3:26 pm

Swapnil Barua wrote:
Wed Apr 21, 2021 8:16 am
You did the summation of infinite series?
No, Telescoping

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