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

**rcs** · March 22nd 2011, 03:54 AM

Please help me on this ... thanks

**Prove It** · March 22nd 2011, 04:22 AM

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

**rcs** · March 22nd 2011, 05:28 AM

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

**Prove It** · March 22nd 2011, 05:41 AM

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

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

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

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

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

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

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

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

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

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

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

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

So your series is

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

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

**rcs** · March 23rd 2011, 03:50 AM

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

**Plato** · March 23rd 2011, 03:59 AM

Originally Posted by rcs

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

Oh no it is not. It is so simple.

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

**Prove It** · March 23rd 2011, 07:15 AM

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...

**rcs** · March 28th 2011, 07:25 AM

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

**HallsofIvy** · March 28th 2011, 07:45 AM

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.

**Prove It** · March 28th 2011, 07:03 PM

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.

**HallsofIvy** · March 29th 2011, 04:25 AM

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

**rcs** · March 29th 2011, 06:07 AM

thanks

honestly i really dont know what telescoping series mean..

**Plato** · March 29th 2011, 06:20 AM

Originally Posted by rcs

honestly i really dont know what telescoping series mean..

Many textbooks call them collapsing sums.
Look at $\left[ {1 - \frac{1}{2}} \right] + \left[ {\frac{1}{2} - \frac{1}{3}} \right] + \left[ {\frac{1}{3} - \frac{1}{4}} \right] + \left[ {\frac{1}{4} - \frac{1}{5}} \right] + \left[ {\frac{1}{5} - \frac{1}<br /> {6}} \right]$ .
Remove the brackets.
Every term subtracts off, collapses, except the first and last.
So we have $\sum\limits_{k = 1}^5 {\left[ {\frac{1}{k} - \frac{1}{{k + 1}}} \right]} = 1 - \frac{1}{6}$

**rcs** · March 29th 2011, 07:16 PM

Thanks Everybody!

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