1. Sigma Notation

$\Sigma [i=1, 15] i(i-1)^2$

2. Originally Posted by yeloc
$\Sigma [i=1, 15] i(i-1)^2$
$\sum_{n=1}^{15}n(n-1)^2=\sum_{n=0}^{15}(n+1)n^2=\sum_{n=1}^{15}n^3+n^ 2$

3. Can you explain how you went from the first step to the second step?

4. Originally Posted by Mathstud28
$\sum_{n=1}^{15}n(n-1)^2=\sum_{n=0}^{15}(n+1)n^2=\sum_{n=1}^{15}n^3+n^ 2$

Originally Posted by yeloc
Can you explain how you went from the first step to the second step?
I can see why you'd be a bit puzzled ..... here's the trick:

Substitute $n - 1 = m \Rightarrow n = m+1$ into $\sum_{n=1}^{15}n(n-1)^2$. Then you get:

$\sum_{m=0}^{{\color{red}14}} (m+1) m^2 = \sum_{m={\color{red}0}}^{14} (m^3+m^2)$.

Note the corrections (in red) to some typos.

5. Please forgive me, but I am still not understanding why you substitute $n=m+1$

6. Originally Posted by yeloc
Please forgive me, but I am still not understanding why you substitute $n=m+1$
Because that's what Mathstud28 did to get the result that you asked about. If you don't like that, just expand the original summand:

$\sum_{i=1}^{15} i (i - 1)^2 = \sum_{i=1}^{15} (i^3 - 2i^2 + i) = \sum_{i=1}^{15} i^3 - 2 \sum_{i=1}^{15} i^2 + \sum_{i=1}^{15}i$.

Now use the usual formulae on each term.

7. Ohh ok...thanks!

8. Originally Posted by mr fantastic
I can see why you'd be a bit puzzled ..... here's the trick:

Substitute $n - 1 = m \Rightarrow n = m+1$ into $\sum_{n=1}^{15}n(n-1)^2$. Then you get:

$\sum_{m=0}^{{\color{red}14}} (m+1) m^2 = \sum_{m={\color{red}0}}^{14} (m^3+m^2)$.

Note the corrections (in red) to some typos.
Thank you for catching that typo...after doing so many infinite sreies you just forget to lower the upper index.