# Thread: Sums of Series

1. ## Sums of Series

Hi the problem that I have is to show $\sum(r^2-r)=kn(n+1)(n-1)$

where $k$ is a rational number. (Also on the sum sign there is $n$ on top and $r=1$ on bottom.

I know what I need to do to start which is show $\sum r^2-\sum r=\frac{1}{6}n(n+1)(2n+1)-\frac{1}{2}n(n+1)$

but my problem is I find it really hard to realise where the common factors are and what I should, if anything expand anywere. I would be really greatful if someone could post a step by step method on how they would do this explaing what has been extracted and what has been expanded. I know its seems alot but I would be really greatful. Thankyou!

2. Originally Posted by NathanBUK
Hi the problem that I have is to show $\sum(r^2-r)=kn(n+1)(n-1)$

where $k$ is a rational number. (Also on the sum sign there is $n$ on top and $r=1$ on bottom.

I know what I need to do to start which is show $\sum r^2-\sum r=\frac{1}{6}n(n+1)(2n+1)-\frac{1}{2}n(n+1)$

but my problem is I find it really hard to realise where the common factors are and what I should, if anything expand anywere. I would be really greatful if someone could post a step by step method on how they would do this explaing what has been extracted and what has been expanded. I know its seems alot but I would be really greatful. Thankyou!
hi

first off , its obvious that n(n+1) is a common factor , and that left you with

n(n+1)[1/6(2n+1)-1/2]

then simplify the expression in the parentheses

n(n+1)(1/3n-1/3)

1/3n(n+1)(n-1)

kn(n+1)(n-1)

where k=1/3

3. Originally Posted by NathanBUK
Hi the problem that I have is to show $\sum(r^2-r)=kn(n+1)(n-1)$

where $k$ is a rational number. (Also on the sum sign there is $n$ on top and $r=1$ on bottom.

I know what I need to do to start which is show $\sum r^2-\sum r=\frac{1}{6}n(n+1)(2n+1)-\frac{1}{2}n(n+1)$

but my problem is I find it really hard to realise where the common factors are and what I should, if anything expand anywere. I would be really greatful if someone could post a step by step method on how they would do this explaing what has been extracted and what has been expanded. I know its seems alot but I would be really greatful. Thankyou!
n(n+1) is a common factor.

4. Originally Posted by mathaddict
hi

first off , its obvious that n(n+1) is a common factor , and that left you with

n(n+1)[1/6(2n+1)-1/3]
thankyou for your help but I dont quite unerstand where -1/3 came from, I must sound so dumb

5. Originally Posted by NathanBUK
thankyou for your help but I dont quite unerstand where -1/3 came from, I must sound so dumb
no worries , my fault . Corrected ..