# Thread: can anybody help me solve the value of this...

1. ## can anybody help me solve the value of this...

2. Have you tried actually evaluating the stuff under the square roots? The first three are perfect squares... I suspect there's a pattern...

3. i solved for the first 3 but will we know it? they only gave the last term after the first 3 terms... too confusing

4. The general term appears to be $\displaystyle \sqrt{1 + \frac{1}{n^2} + \frac{1}{(n+1)^2}}$

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

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

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

$\displaystyle = \sqrt{\frac{n^4 + 2n^3 + n^2 + 2n^2 + 2n + 1}{n^2(n+1)^2}}$

$\displaystyle = \sqrt{\frac{n^4 + 2n^3 + 3n^2 + 2n + 1}{n^2(n+1)^2}}$

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

$\displaystyle = \frac{n^2 + n + 1}{n(n + 1)}$

$\displaystyle = \frac{n^2 + n + 1}{n^2 + n}$

$\displaystyle = 1 + \frac{1}{n^2 + n}$

$\displaystyle = 1 + \frac{1}{n(n + 1)}$

$\displaystyle = 1 + \frac{1}{n} - \frac{1}{n + 1}$.

$\displaystyle \sum_{n = 1}^{1999}\left(1 + \frac{1}{n} - \frac{1}{n + 1}\right)$.

This is a telescopic series... See how you go evaluating it...

5. too long to be manipulated by summation... thanks anyway

6. Originally Posted by rcs
too long to be manipulated by summation... thanks anyway
Oh no it is not. It is so simple.

$\displaystyle \sum_{n = 1}^{1999}\left(1 + \frac{1}{n} - \frac{1}{n + 1}\right)$= $\displaystyle\sum_{n = 1}^{1999}1+\sum_{n = 1}^{1999}\left(\frac{1}{n} - \frac{1}{n + 1} \right)$= $1999+1-\frac{1}{2000}$

7. Originally Posted by rcs
too long to be manipulated by summation... thanks anyway
I see - well next time you post a difficult looking problem I'll just say it'll take up too much of my time, considering you're not even going to put in any effort...

8. i would like to apologize to what i had replied on Prove it's answer... i never meant to be so mean... my point was, i didn't understand what he had been solving, i mean the answer itself is too difficult for me to understand , i was thinking that i would be solving too long like adding from 1 to 1999 consecutive numbers because it is summation ... i didn't know how it would go to 1999 + 1 - 1/2000 .... sorry i really cannot get how it would become like that... sorry hope you guys understand... it was just miscommunication.

thanks

9. Do you know what "telescoping series" means?

Prove It showed that the series can be written
$\displaytype \sum_{n=1}^{1999}\left(1+ \frac{1}{n}- \frac{1}{n+1}\right)$

That is,
$(1+ 1- \frac{1}{2})+ (1+ \frac{1}{2}- \frac{1}{3})+ (1+ \frac{1}{3}+ \frac{1}{4}+ (1+ \frac{1}{4}- \frac{1}{5}) \cdot\cdot\cdot$

Do you see how the fractions cancel? All except in the last term, $-\frac{1}{2000}$. Of course, you will also have "1" added 1999 times.

10. Originally Posted by HallsofIvy
Do you know what "telescoping series" means?

Prove It showed that the series can be written
$\displaytype \sum_{n=1}^{1999}\left(1+ \frac{1}{n}- \frac{1}{n+1}\right)$

That is,
$(1+ 1- \frac{1}{2})+ (1+ \frac{1}{2}- \frac{1}{3})+ (1+ \frac{1}{3}+ \frac{1}{4}+ (1+ \frac{1}{4}- \frac{1}{5}) \cdot\cdot\cdot$

Do you see how the fractions cancel? All except in the last term, $-\frac{1}{2000}$. Of course, you will also have "1" added 1999 times.
I'm sure you mean

$\displaystyle \left(1 + 1 - \frac{1}{2}\right) + \left(1 + \frac{1}{2} - \frac{1}{3}\right) + \left(1 + \frac{1}{3} - \frac{1}{4}\right) + \left(1 + \frac{1}{4} - \frac{1}{5}\right) + \dots + \left(1 + \frac{1}{1999} - \frac{1}{2000}\right)$.

But yes, a telescopic series means that all terms except for the first and last end up cancelling.

11. Yes, thanks. I accidently changed a "-" to a "+".

12. thanks

honestly i really dont know what telescoping series mean..

13. Originally Posted by rcs
honestly i really dont know what telescoping series mean..
Many textbooks call them collapsing sums.
Look at $\left[ {1 - \frac{1}{2}} \right] + \left[ {\frac{1}{2} - \frac{1}{3}} \right] + \left[ {\frac{1}{3} - \frac{1}{4}} \right] + \left[ {\frac{1}{4} - \frac{1}{5}} \right] + \left[ {\frac{1}{5} - \frac{1}
{6}} \right]$
.
Remove the brackets.
Every term subtracts off, collapses, except the first and last.
So we have $\sum\limits_{k = 1}^5 {\left[ {\frac{1}{k} - \frac{1}{{k + 1}}} \right]} = 1 - \frac{1}{6}$

14. Thanks Everybody!