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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

Community
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AncientSwordRage
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  • It's good that you have edited out the minor mistakes in your equivalent expression, but by doing so without noting that you've done so you've made the first part of my answer look somewhat ridiculous. – Gerry Myerson Mar 20 '12 at 22:40
  • Sorry @GerryMyerson the same thing is usually done on the other stackexchange sites I frequent with little complaint. I'll add a note into my question. – AncientSwordRage Mar 21 '12 at 08:01
  • @Pureferret: I for one always complain when I see it done. It makes it look as if someone else made the mistakes. – joriki Mar 21 '12 at 08:14
  • @joriki that's fair enough, but if I ever see it, I check what was changed in the last edit, assuming the advice given was followed. I'll try to remember next time it happens here. – AncientSwordRage Mar 21 '12 at 08:16

2 Answers2

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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. .

Gerry Myerson
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  • Gerry, that's very helpful but I can't find a satisfactory explanation for resulting formula. All I found was this quote: "The sum of squares identity is a different matter. There are only two kinds of proofs for this identity in calculus textbooks, one using induction and the other involving a telescoping sum of cubes. I couldn't find a single textbook that gave a proof different from these two. Neither of these gives much of a feel for why the formula looks like it does." The best ways I've seen are to geometrically construct shapes or diagrams and find a pattern form counting. – AncientSwordRage Mar 20 '12 at 12:19
  • But I'm after something more like Gauss' proof of the sum of integers. – AncientSwordRage Mar 20 '12 at 12:20
  • So you are changing your question from "how can I evaluate this sum?" to "find me a proof that I will like," is that right? Sorry, I don't know if there is one. – Gerry Myerson Mar 20 '12 at 12:35
  • No, I don't think I should change it, I might ask another question though. – AncientSwordRage Mar 20 '12 at 15:33
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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 $$\begin{eqnarray}\sum_{x=0}^{n+1} x^\underline{a} & = & (n+1)^\underline{a} + \sum_{x=0}^{n} x^\underline{a} \\ & = & (n+1)^\underline{a} + \frac{(n+1)^\underline{a+1}}{a+1} \\ & = & \frac{(a+1)(n+1)^\underline{a} + (n+1)^\underline{a+1}}{a+1} \\ & = & \frac{(a+1)(n+1)^\underline{a} + (n+1)^\underline{a}(n-a+1)}{a+1} \\ & = & \frac{(n+2)(n+1)^\underline{a}}{a+1} \\ & = & \frac{(n+2)^\underline{a+1}}{a+1} \end{eqnarray}$$

Peter Taylor
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