Sum of Series

**cedricc** · April 25th 2010, 12:56 AM

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??

**BabyMilo** · April 25th 2010, 02:51 AM

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?

**cedricc** · April 25th 2010, 02:54 AM

i don't really follow, sorry...

**BabyMilo** · April 25th 2010, 03:01 AM

Originally Posted by cedricc

i don't really follow, sorry...

which part?

**cedricc** · April 25th 2010, 03:12 AM

Actually, nevermind, i got it ^^ Thanks!

**Wilmer** · April 25th 2010, 03:56 AM

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

**Krizalid** · April 25th 2010, 01:33 PM

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.

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