# Thread: Trying to find the sum of a series

1. ## Trying to find the sum of a series

Hello I have found a pretty complicated way of finding the sum to this series, and was wondering if anyone could help me to find a simpler way.

The sum of the series 1+2+3+4+....+n= n(n+1)/2

What is the sum of 2+6+20+....+n(n+1)=???

2. Originally Posted by Finney
Hello I have found a pretty complicated way of finding the sum to this series, and was wondering if anyone could help me to find a simpler way.

The sum of the series 1+2+3+4+....+n= n(n+1)/2

What is the sum of 2+6+20+....+n(n+1)=???
I was wondering is the sum of series you are looking for actually:
2+6+12+20+....+n(n+1)=?

Because it's:
1(2)+2(3)+3(4)+4(5)+...+n(n+1)

3. If you are looking for sum of series:
$2 + 6 + 12 + 20 + .... + n(n + 1) =?$

then its:
$2 + 6 + 12 + 20 + .... + n(n + 1) = \frac{{n(n + 1)(n + 2)}}{3}$

You need to look for:
$2(1 + 3 + 6 + 10 + ... + \frac{{n(n + 1)}}{2}) = 2(\frac{{n(n + 1)(n + 2)}}{6}) = \frac{{n(n + 1)(n + 2)}}{3}$

Look at this post:
http://www.mathhelpforum.com/math-he...1-2-3-4-a.html

4. Originally Posted by Finney
Hello I have found a pretty complicated way of finding the sum to this series, and was wondering if anyone could help me to find a simpler way.

The sum of the series 1+2+3+4+....+n= n(n+1)/2

What is the sum of 2+6+20+....+n(n+1)=???
You have,
$\sum_{k=1}^n k(k+1)=$ $\sum_{k=1}^n k^2+k=$ $\sum_{k=1}^n k^2+\sum_{k=1}^n k$
Using some theorems,
$\frac{n(2n+1)(n+1)}{6}+\frac{n(n+1)}{2}$

5. ## Aha

Thats very clever, the above post method I tried but it did not fork by far.