# Using an identity to find the sum to n terms of a series

1. Dec 2, 2005

### mr bob

Just working through my FP1 book and have got stuck on a question.
Use the identity $(r+1)^3 - r^3 \equiv3r^2 + 3r + 1$
to find $\sum\limits_{r = 1}^n r(r+1)$
I've tried using the method of differences to get $n^3 + 3n^2 + 3n$, but cant see how to get it back into its original form, not sure how the identity corresponds to r(r+1).

Last edited: Dec 2, 2005
2. Dec 2, 2005

### hypermonkey2

at a glance, i would say that you need to find a telescoping series somewhere.

3. Dec 2, 2005

### hypermonkey2

wait, so youre not allowed to simplify it to the sum of r^2 + r? because then you change just seperate it to the sum of the first n r^2 plus the first n of r. which should give (n)(n+1)(2n+1)/6 + (n)(n+1)/2.
this is what i see.

4. Dec 2, 2005

### mr bob

I could simplify it, but the question asks to use the identity. Im not sure how to use that with that series. Although splitting it down into its standard results would be alot easier.

5. Dec 2, 2005

### Fermat

rearrange the identity like this,

3r(r+1) = (r+1)^3 - r^3 - 1

Then

Sigma r(r+1) = (1/3) Sigma {(r+1)^3 - r^3 - 1}

Now use the http://thesaurus.maths.org/mmkb/entry.html?action=entryById&id=3935" [Broken] in the rhs (right hand side) and simplify.

It would be a lot simpler doing it the other way though, like hypermonkey suggested.

Last edited by a moderator: May 2, 2017
6. Dec 2, 2005

### Hurkyl

Staff Emeritus
Why would you ever want to use the sum of cubes formula for that sum?

7. Dec 2, 2005

### Werg22

Alright I made it shorter; so here is what you need to show

$$\sum_{k=1}^{n-1} (k + 1)^{3} - k^{3} = \sum_{k=1}^{n-1} 3k^{2} + 3k + 1$$

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

$$\sum_{k=1}^{n-1} 3k^{2} + 3k + 1 = n^{3} - 1$$

$$3\sum_{k=1}^{n-1}k^{2} + 3\sum_{k=1}^{n-1}k + n - 1 = n^{3} - 1$$

$$; 3\sum_{k=1}^{n-1}k = \frac{3n(n-1)}{2}$$

$$3\sum_{k=1}^{n-1}k^{2} + \frac{3n(n-1)}{2}= n^{3} - n$$

$$3\sum_{k=1}^{n}k^{2} + \frac{3(n+1)(n)}{2}= (n+1)^{3} - (n+1)$$

$$3\sum_{k=1}^{n}k^{2} = \frac{n(n+1)(2n + 1)}{2}$$

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

Then knowing that $$\sum_{k=1}^{n}k = \frac{n(n+1)}{2}$$

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

Last edited: Dec 2, 2005
8. Dec 3, 2005

### Fermat

This simplifies down a bit further to,

$$\sum_{k=1}^{n}k^{2} + k = \frac{n(n+1)(n + 2)}{3}$$