A problem about the ratio of atoms I'm trying to find the ratio of the number of atoms in the core of a nanoparticle, and the outermost shell with each shell containing:
$10k^2+2$ atoms
Then the total number up to the $k^{th}$ shell is (I believe)
$\sum\limits_{n=1}^{k-1}(10(k-n)^2+2)$
It's this expression I'm stuck on. I'm not 100% on what I'm allowed to do to evaluate this summation. I couldn't find much about operations on summations when I went looking.
Is this expression equivalent?
$10\bigg(\sum\limits_{n=1}^{k-1}k^2+\sum\limits_{n=1}^{k-1}n^2-\sum\limits_{n=1}^{k-1}2kn\bigg)+\sum\limits_{n=1}^{k-1}2$
If so how can I evaluate these?
Edit: Note, errors fixed thanks to Gerry's advice in his answer
 A: Remember $(k-n)^2=k^2+n^2-2kn$, so you're missing a factor of 2. Also, the 10 doesn't multiply the +2 in the sum, so your last sum shouldn't be in the parentheses. 
But you are making it a bit harder than it has to be. Do you see that $$\sum_{n=1}^{k-1}(10(k-n)^2+2)=\sum_{n=1}^{k-1}(10n^2+2)$$ On the left, as $n$ goes from $1$ to $k-1$, $k-n$ goes from $k-1$ to $1$, so the form on the right adds up the same numbers. 
Now all you need is the formula for $\sum_{n=1}^{k-1}n^2$, which you can find in any number of places. . 
A: The rearrangement is fine. You can always reorder and regroup terms in a finite summation - although with infinite summations you do need to be careful.
To evaluate the sums, one useful trick to learn is to rearrange them into sums of falling powers $x^\underline{a} = x(x-1)\ldots(x-a+1)$ and then use $$\sum_{x=0}^n x^\underline{a} = \frac{(n+1)^\underline{a+1}}{a+1}$$ This is a discrete analogue to $$\int_0^n x^a dx = \frac{n^{a+1}}{a+1}$$ which might help your intuition. Proof is by induction:
First note that if $x,a\in\mathbb{N}$ and $a > x$ then $x^\underline{a} = 0$ because it's a product which includes a $0$. So for base cases $n=0$ to $n=a-1$ we have $\textrm{LHS} = \textrm{RHS} = 0$.
Induction step: assume the identity holds for $n$; therefore $$\begin{eqnarray}\sum_{x=0}^{n+1} x^\underline{a} & = & (n+1)^\underline{a} + \sum_{x=0}^{n} x^\underline{a} \\
& = & (n+1)^\underline{a} + \frac{(n+1)^\underline{a+1}}{a+1} \\
& = & \frac{(a+1)(n+1)^\underline{a} + (n+1)^\underline{a+1}}{a+1} \\
& = & \frac{(a+1)(n+1)^\underline{a} + (n+1)^\underline{a}(n-a+1)}{a+1} \\
& = & \frac{(n+2)(n+1)^\underline{a}}{a+1} \\
& = & \frac{(n+2)^\underline{a+1}}{a+1} \end{eqnarray}$$
