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

[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 can't see how to get it back into its original form, not sure how the identity corresponds to r(r+1).

 

[hypermonkey2]

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

 

[hypermonkey2]

wait, so youre not allowed to simplify it to the sum of r^2 + r? because then you change just separate 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.

 

[mr bob]

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

 

[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" in the rhs (right hand side) and simplify.

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

 

[Hurkyl]

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

 

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

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

 

[Fermat]

This simplifies down a bit further to,

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