# Thread: Sum proof...need a little help

1. ## Sum proof...need a little help

$\displaystyle \sum_{i=1}^{n-1} i(i+1) = (n(n-1)(n+1))/3.$
For all integers $\displaystyle n \geq 2$

Base case: lets n = 2

$\displaystyle \sum_{i=1}^{n-1} i(i+1) = (2(2-1)(2+1))/3.$

= 2(1)(3)
=2

Inductive step:

$\displaystyle \sum_{i=1}^{(n+1)-1} i(i+1) =$$\displaystyle \sum_{i=1}^{n} i(i+1) + [(n+1)((n+1)+1)]$

$\displaystyle (n(n-1)(n+1))/3 + [(n+1)(n+2)]$

$\displaystyle (n(n-1)(n+1)+3(n+1)(n+2))/3$

$\displaystyle ((n+1)(n^2+2n+6))/3$

And this is where i am stuck cause I dont really know what to do with the
$\displaystyle n^2+2n+6.$

2. Take note of the special sums:
$\displaystyle \sum\limits_{k = 1}^n k = \frac{{n(n + 1)}}{2}\;\& \,\sum\limits_{k = 1}^n {k^2 } = \frac{{n(n + 1)(2n + 1)}}{6}$

Then write $\displaystyle \sum\limits_{k = 1}^n {k(k + 1)} = \sum\limits_{k = 1}^n {k^2 } +\sum\limits_{k = 1}^n k$.

3. Originally Posted by Plato
Take note of the special sums:
$\displaystyle \sum\limits_{k = 1}^n k = \frac{{n(n + 1)}}{2}\;\& \,\sum\limits_{k = 1}^n {k^2 } = \frac{{n(n + 1)(2n + 1)}}{6}$

Then write $\displaystyle \sum\limits_{k = 1}^n {k(k + 1)} = \sum\limits_{k = 1}^n {k^2 } +\sum\limits_{k = 1}^n k$.

I just substutited $\displaystyle \sum\limits_{k = 1}^n {k(k + 1)}$
by the given statement. I dunno how those special cases would come into play for this..

4. Originally Posted by Kitizhi
I dunno how those special cases would come into play for this..
How much do you know about sums?
$\displaystyle \begin{gathered} \sum\limits_k {\left( {a_k + b_k } \right)} = \sum\limits_k {\left( {a_k } \right)} + \sum\limits_k {\left( {b_k } \right)} \hfill \\ \sum\limits_k {k\left( {k + 1} \right)} = \sum\limits_k {\left( {k^2 + k} \right)} = \sum\limits_k {\left( {k^2 } \right)} + \sum\limits_k {\left( k \right)} \hfill \\ \end{gathered}$

Using induction is a waste of resources.
If you are asked to study SUMS then you ought to know about SPECIAL SUMS.

5. not that much just some of the basic rules, no special cases.

I have this answer to this question and we mainly focus on the use of induction but i dont fully understand it which is why I am seeking help.

6. Originally Posted by Kitizhi
not that much just some of the basic rules, no special cases.
I have this answer to this question and we mainly focus on the use of induction but i dont fully understand it which is why I am seeking help.
Well good luck with that.
I do think it is a pity that you are being asked to reinvent a wheel.