i need to find the sum of :
I'm getting the wrong answer, just cant find out where i'm going wrong.
here's what i did...
replaced the above by;
& the given answer is 35/429
Hello, adhyeta!
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] $Find the sum: .$\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: $\displaystyle \frac{35}{429}$
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] $
. .$\displaystyle \frac{1}{7\cdot9\cdot11} \;=\;\frac{1}{8}\bigg[\frac{{\color{red}\rlap{/}}1}{{\color{red}\rlap{/}}7} - \frac{{\color{green}\rlap{/}}2}{{\color{green}\rlap{/}}9} + \frac{1}{11}\bigg]$
. $\displaystyle \frac{1}{9\cdot11\cdot13} \;=\;\frac{1}{8}\bigg[\frac{{\color{green}\rlap{/}}1}{{\color{green}\rlap{/}}9} - \frac{2}{11} + \frac{1}{13}\bigg]$
Add the equations:
. . $\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}} $
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 $\displaystyle \frac{1}{4}\left( \frac{1}{3}-\frac{1}{11\times 13} \right)=\frac{1}{4}\times \frac{140}{429}=\frac{35}{429}.$
Hello, adhyeta!
At this point, I'd factor again . . .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;
$\displaystyle \sum_{n=1}^{5}\frac{1}{(2n-1)(2n+1)(2n+3)}$
. . $\displaystyle =\; \frac{1}{2}\sum_{n=1}^{5}\frac{(2n+3)-(2n+1)}{(2n-1)(2n+1)(2n+3)}$
. . $\displaystyle =\; \frac{1}{2}\sum_{n=1}^{5}\left[\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n-1)(2n+3)}\right]$
. . $\displaystyle =\; \frac{1}{2}\sum_{n=1}^{5}\left[\frac{1}{2}\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}-\frac{1}{4}\frac{(2n+3)-(2n-1)}{(2n-1)(2n+3)}\right]$
. . $\displaystyle =\;\frac{1}{4}\sum_{n=1}^{5}\left[\frac{1}{(2n-1)}-\frac{1}{(2n+1)}-\frac{1}{2}\frac{1}{(2n-1)}+\frac{1}{2}\frac{1}{(2n+3)}\right]$
$\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 . . .