<s>[INST] My son's Sum of Some is beautiful! But what is the proof or explanation? My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation:
$$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$ 
Why is this so? What is the proof or explanation? Is it true for any $n$?
His own presentation of his finding:

Every one of these numbers is two times the number before it.
  $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.
  I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.
  If I then skip three ones and read the fourth, that one equals my sum!

 [/INST] Here is another formula similar to your son's, using powers of $5$:
$$1000000\cdot5^n+100000\cdot5^{n+1}+10000\cdot5^{n+2}+1000\cdot5^{n+3}+100\cdot5^{n+4}+5^{n+6}=5^{n+9}$$
We can derive identities of a similar form more generally.  Consider any $m$th degree polynomial $f(x)\in\mathbb{Z}[x]$ of the following form
$$f(x)=a_0+a_1x+\ldots+a_{m-1}x^{m-1}-x^m$$
Notice that all rational roots of $f(x)$ are integers.  Suppose $f$ has a rational root $k$.  Then we have the following equality:
$$\begin{align}k^{n+m}&=k^nk^m\\&=k^n(a_0+a_1k+a_2k^2+\ldots+a_{m-1}k^{m-1})\\&=a_0k^n+a_1k^{n+1}+a_2k^{n+2}+\ldots+a_{m-1}k^{n+m-1}\end{align}$$
Your son's formula involves the $7$th degree polynomial $f(x)=100+10x+x^3-x^7$ and the root $k=2$.  Above I've used the polynomial $f(x)=10^6+10^5x+10^4x^2+10^3x^3+10^2x^4+x^6-x^9$ and the root $k=5$.
Here are other examples for $k=3$.  Consider the polynomials $f(x)=3+8x+80x^3-x^7,$ $g(x)=2130+10x+x^3-x^7$, and $h(x)=99+687x+x^3-x^7$, and notice that $f(3)=g(3)=h(3)=0$.  These give us the following identities:
$$3\cdot3^n+8\cdot 3^{n+1}+80\cdot3^{n+3}=3^{n+7}\\2130\cdot3^n+10\cdot3^{n+1}+3^{n+3}=3^{n+7}\\99\cdot3^n+687\cdot3^{n+1}+3^{n+3}=3^{n+7}$$
Try using this method to find identities for other values of $k$ (including negative integers!).
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