# Thread: Sums of Series

1. ## Sums of Series

Hi, well got my exam tomorrow and this is the only topic area I struggle with soo much, a past paper i'm trying here asks this qustion:

It is given that $Sn=\sum(3r^2-3r+1)$

Use the formulae for $\sum(r^2)$ and $\sum(r)$ to show that $Sn=n^3$.

I know that I treat the +1 as n but I really dont know where I should take out factors or when I should expand etc. Also I get the formulas for the r squared and r so im aware of what they are.

Each time Ive tried Ive managed to get to either $n^3-1$ or $n(n^2+1)$

I would be really greatful if someone could explain what process I should do to get the answer.

Thankyou !

2. Originally Posted by NathanBUK
Hi, well got my exam tomorrow and this is the only topic area I struggle with soo much, a past paper i'm trying here asks this qustion:

It is given that $Sn=\sum(3r^2-3r+1)$

Use the formulae for $\sum(r^2)$ and $\sum(r)$ to show that $Sn=n^3$.

I know that I treat the +1 as n but I really dont know where I should take out factors or when I should expand etc. Also I get the formulas for the r squared and r so im aware of what they are.

Each time Ive tried Ive managed to get to either $n^3-1$ or $n(n^2+1)$

I would be really greatful if someone could explain what process I should do to get the answer.

Thankyou !
You probably went about it right but made some small error somewhere along the way. Here are my steps.

$\sum_{r=1}^nr^2=\frac{n(n+1)(2n+1)}{6}$

$\sum_{r=1}^nr=\frac{n(n+1)}{2}$

So

$\sum_{r=1}^n\left(3r^2-3r+1\right)= \sum_{r=1}^n3r^2-\sum_{r=1}^n3r+\sum_{r=1}^n1 =$

$3\sum_{r=1}^nr^2-3\sum_{r=1}^nr+\sum_{r=1}^n1 = 3\left(\frac{n(n+1)(2n+1)}{6}\right)-3\left(\frac{n(n+1)}{2}\right)+n =$

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

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

$n(n+1)(n-1) + n = n(n^2-1) + n = n(n^2 - 1 + 1) = n(n^2) = n^3$

3. Ohh thankyou ! Lets just hope one too nasty doesnt come up on my test