# Math Help - Sum of Series

1. ## Sum of Series

Find the sum of the first n terms of the series:

(i) 1.3.5 + 3.5.7 + 5.7.9 + ... + (2n - 1)(2n + 1)(2n +3)

(ii) 1/(1.3.5) + 1/(3.5.7) + 1/(5.7.9) + ... + 1/[(2n - 1)(2n + 1)(2n +3)]

I have tried many methods, but with no avail... how can i solve this while displaying working and explanations??

2. Originally Posted by cedricc
Find the sum of the first n terms of the series:

(i) 1.3.5 + 3.5.7 + 5.7.9 + ... + (2n - 1)(2n + 1)(2n +3)

(ii) 1/(1.3.5) + 1/(3.5.7) + 1/(5.7.9) + ... + 1/[(2n - 1)(2n + 1)(2n +3)]

I have tried many methods, but with no avail... how can i solve this while displaying working and explanations??
i duno how to do the sum sign, so im sorry. so when i do the sum sign that means r=1 to n.

but i) $(2n - 1)(2n + 1)(2n +3)$ times that out.
which comes to $8n^3+12n^2-2n-3$.

so for the sum from r=1 to n = $8\sum r^3 + 12\sum r^2 -2 \sum r -3r$.

then using the fact

$\sum r= \frac{1}{2}n(n+1)$
$\sum r^2= \frac{1}{6}n(n+1)(2n+1)$
$\sum r^3= \frac{1}{4}n^2(n+1)^2$

sub this in $8\sum r^3 + 12\sum r^2 -2 \sum r -3r$.

do you understand?

3. i don't really follow, sorry...

4. Originally Posted by cedricc
i don't really follow, sorry...
which part?

5. Actually, nevermind, i got it ^^ Thanks!

6. Go here: The On-Line Encyclopedia of Integer Sequences
Enter A061550 in search box

7. Originally Posted by cedricc
(ii) 1/(1.3.5) + 1/(3.5.7) + 1/(5.7.9) + ... + 1/[(2n - 1)(2n + 1)(2n +3)]
this part is much interesting, it can be solved in few lines:

$\frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{1}{4}\cdot \frac{(2n+3)-(2n-1)}{(2n-1)(2n+1)(2n+3)},$ thus the general term equals $\frac{1}{4}\left( \frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)} \right).$ Now let $a_n=\frac{1}{(2n-1)(2n+1)}$ and see that $a_{n+1}=\frac{1}{(2n+1)(2n+3)},$ so $\frac{1}{4}\sum\limits_{n=1}^{m}{\left( \frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)} \right)}$ is a telescoping sum.