# Thread: sum of telescoping series...

1. ## sum of telescoping series...

i need to find the sum of :

$\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+\frac{1}{7.9.11}+\frac{1}{9.11.13}$

I'm getting the wrong answer, just cant find out where i'm going wrong.

here's what i did...

replaced the above by;

$\sum_{1}^{5}\frac{1}{(2n-1)(2n+1)(2n+3)}$

$1/2\sum_{1}^{5}\frac{(2n+3)-(2n+1)}{(2n-1)(2n+1)(2n+3)}$

$1/2(\sum_{1}^{5}\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n-1)(2n+3)})$

$1/2(\sum_{1}^{5}1/2\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}-1/4\frac{(2n+3)-(2n-1)}{(2n-1)(2n+3)})$

$1/4(\sum_{1}^{5}\frac{1}{(2n-1)}-\frac{1}{(2n+1)}-1/2\frac{1}{(2n-1)}+1/2\frac{1}{(2n+3)})$

$1/4(\sum_{1}^{5}(1-1/11)-1/2(1-1/13))$

$1/4(\sum_{1}^{5}(10/11)-1/2(12/13))=16/143$

& the given answer is 35/429

i need to find the sum of :

$\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+\frac{1}{7.9.11}+\frac{1}{9.11.13}$

I'm getting the wrong answer, just cant find out where i'm going wrong.

here's what i did...

replaced the above by;

$\sum_{1}^{5}\frac{1}{(2n-1)(2n+1)(2n+3)}$

$1/2\sum_{1}^{5}\frac{(2n+3)-(2n+1)}{(2n-1)(2n+1)(2n+3)}$

$1/2(\sum_{1}^{5}\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n-1)(2n+3)})$

$1/2(\sum_{1}^{5}1/2\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}-1/4\frac{(2n+3)-(2n-1)}{(2n-1)(2n+3)})$

$1/4(\sum_{1}^{5}\frac{1}{(2n-1)}-\frac{1}{(2n+1)}-1/2\frac{1}{(2n-1)}+1/2\frac{1}{(2n+3)})$

$1/4(\sum_{1}^{5}(1-1/11)-1/2(1-1/13))$

$1/4(\sum_{1}^{5}(10/11)-1/2(12/13))=16/143$

& the given answer is 35/429
you could split it up further

$\displaystyle \frac{1}{(2n-1)(2n+1)(2n+3)} = \frac{1}{8} \left( \frac{1}{2n-1} - \frac{1}{2n+1} + \frac{1}{2n+3}\right)$

Find the sum: .$\displaystyle S \;=\;\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot 7}+\frac{1}{5\cdot7\cdot9}+\frac{1}{7\cdot9\cdot11 }+\frac{1}{9\cdot11\cdot13}$

The given answer is: $\displaystyle \frac{35}{429}$
I would make three fractions: . $\displaystyle \frac{1}{(2n-1)(2n+1)(2n+3)} \;=\;\frac{1}{8}\,\bigg[\frac{1}{2n-1} - \frac{2}{2n+1} + \frac{1}{2n+3}\bigg]$

Then we have:

. . $\displaystyle \frac{1}{1\cdot3\cdot5} \;=\; \frac{1}{8}\left[\frac{1}{1} - \frac{2}{3} + \frac{{\color{cyan}\rlap{/}}1}{{\color{cyan}\rlap{/}}5}\right]$

. . $\displaystyle \frac{1}{3\cdot5\cdot7} \;=\;\frac{1}{8}\bigg[\frac{1}{3} - \frac{{\color{cyan}\rlap{/}}2}{{\color{cyan}\rlap{/}}5} + \frac{{\color{red}\rlap{/}}1}{{\color{red}\rlap{/}}7}\bigg]$

. . $\displaystyle \frac{1}{5\cdot7\cdot9} \;=\;\frac{1}{8}\bigg[\frac{{\color{cyan}\rlap{/}}1}{{\color{cyan}\rlap{/}}5} - \frac{{\color{red}\rlap{/}}2}{{\color{red}\rlap{/}}7} + \frac{{\color{green}\rlap{/}}1}{{\color{green}\rlap{/}}9}\bigg]$

. .$\displaystyle \frac{1}{7\cdot9\cdot11} \;=\;\frac{1}{8}\bigg[\frac{{\color{red}\rlap{/}}1}{{\color{red}\rlap{/}}7} - \frac{{\color{green}\rlap{/}}2}{{\color{green}\rlap{/}}9} + \frac{1}{11}\bigg]$

. $\displaystyle \frac{1}{9\cdot11\cdot13} \;=\;\frac{1}{8}\bigg[\frac{{\color{green}\rlap{/}}1}{{\color{green}\rlap{/}}9} - \frac{2}{11} + \frac{1}{13}\bigg]$

. . $\displaystyle S \;=\;\frac{1}{8}\bigg[1 - \frac{2}{3} + \frac{1}{3} + \frac{1}{11} - \frac{2}{11} + \frac{1}{13}\bigg] \;=\;\frac{1}{8}\cdot \frac{280}{429} \;=\;\boxed{\frac{35}{429}}$

4. Originally Posted by Soroban

I would make three fractions: . $\displaystyle \frac{1}{(2n-1)(2n+1)(2n+3)} \;=\;\frac{1}{8}\,\bigg[\frac{1}{2n-1} - \frac{2}{2n+1} + \frac{1}{2n+3}\bigg]$

Then we have:

. . $\displaystyle \frac{1}{1\cdot3\cdot5} \;=\; \frac{1}{8}\left[\frac{1}{1} - \frac{2}{3} + \frac{{\color{cyan}\rlap{/}}1}{{\color{cyan}\rlap{/}}5}\right]$

. . $\displaystyle \frac{1}{3\cdot5\cdot7} \;=\;\frac{1}{8}\bigg[\frac{1}{3} - \frac{{\color{cyan}\rlap{/}}2}{{\color{cyan}\rlap{/}}5} + \frac{{\color{red}\rlap{/}}1}{{\color{red}\rlap{/}}7}\bigg]$

. . $\displaystyle \frac{1}{5\cdot7\cdot9} \;=\;\frac{1}{8}\bigg[\frac{{\color{cyan}\rlap{/}}1}{{\color{cyan}\rlap{/}}5} - \frac{{\color{red}\rlap{/}}2}{{\color{red}\rlap{/}}7} + \frac{{\color{green}\rlap{/}}1}{{\color{green}\rlap{/}}9}\bigg]$

. .$\displaystyle \frac{1}{7\cdot9\cdot11} \;=\;\frac{1}{8}\bigg[\frac{{\color{red}\rlap{/}}1}{{\color{red}\rlap{/}}7} - \frac{{\color{green}\rlap{/}}2}{{\color{green}\rlap{/}}9} + \frac{1}{11}\bigg]$

. $\displaystyle \frac{1}{9\cdot11\cdot13} \;=\;\frac{1}{8}\bigg[\frac{{\color{green}\rlap{/}}1}{{\color{green}\rlap{/}}9} - \frac{2}{11} + \frac{1}{13}\bigg]$

. . $\displaystyle S \;=\;\frac{1}{8}\bigg[1 - \frac{2}{3} + \frac{1}{3} + \frac{1}{11} - \frac{2}{11} + \frac{1}{13}\bigg] \;=\;\frac{1}{8}\cdot \frac{280}{429} \;=\;\boxed{\frac{35}{429}}$

hey soroban. i got that.
thnx

but what do you think might be wrong with my solution.
i have four terms you have combined 2 of them as one & then solved.

but doing it the other way should also give the same answer, right?
what do you think might be wrong with it?

5. we have $\displaystyle \frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{1}{4}\left( \frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)} \right),$ thus from here we've automatically found a telescoping sum, thus your sum equals $\displaystyle \frac{1}{4}\left( \frac{1}{3}-\frac{1}{11\times 13} \right)=\frac{1}{4}\times \frac{140}{429}=\frac{35}{429}.$

Find the sum:

$\displaystyle \frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\f rac{1}{5\cdot7\cdot9}+\frac{1}{7\cdot9\cdot11}+\fr ac{1}{9\cdot1\cdot.13}$

I'm getting the wrong answer, just cant find out where i'm going wrong.

here's what i did...

replaced the above by;

$\displaystyle \sum_{n=1}^{5}\frac{1}{(2n-1)(2n+1)(2n+3)}$

. . $\displaystyle =\; \frac{1}{2}\sum_{n=1}^{5}\frac{(2n+3)-(2n+1)}{(2n-1)(2n+1)(2n+3)}$

. . $\displaystyle =\; \frac{1}{2}\sum_{n=1}^{5}\left[\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n-1)(2n+3)}\right]$

. . $\displaystyle =\; \frac{1}{2}\sum_{n=1}^{5}\left[\frac{1}{2}\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}-\frac{1}{4}\frac{(2n+3)-(2n-1)}{(2n-1)(2n+3)}\right]$

. . $\displaystyle =\;\frac{1}{4}\sum_{n=1}^{5}\left[\frac{1}{(2n-1)}-\frac{1}{(2n+1)}-\frac{1}{2}\frac{1}{(2n-1)}+\frac{1}{2}\frac{1}{(2n+3)}\right]$
At this point, I'd factor again . . .

$\displaystyle \frac{1}{8}\sum^5_{n=1}\left[\frac{2}{2n-1} - \frac{2}{2n+1} - \frac{1}{2n-1} + \frac{1}{2n+3}\right]$

. . $\displaystyle = \; \frac{1}{8}\sum^5_{n=1}\bigg[\frac{1}{2n-1} - \frac{2}{2n+1} + \frac{1}{2n+3}\bigg]$

And this is equivalent to what I had found . . .