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!


The result simply holds because 
$$\begin{align}
(100*2^n)+(10*2^{n+1})+2^{n+3} &=(2^6+2^5+2^2)*2^n+(2^3+2)*2^{n+1}+2^{n+3} \\
& = 2^{n+6} +2^{n+5}+2^{n+2} +2^{n+4} + 2^{n+2}+2^{n+3} \\
& =2^{n+6}+2^{n+5}+2^{n+4} + 2^{n+3}+(2^{n+2}+2^{n+2})\\
& =2^{n+6}+2^{n+5}+2^{n+4}  +(2^{n+3}+2^{n+3})\\
& =2^{n+6}+2^{n+5}+(2^{n+4}+2^{n+4})\\
& =2^{n+6}+(2^{n+5}+2^{n+5})\\
& =(2^{n+6}+2^{n+6})\\
&=\mathbf{2^{n+7}}
\end{align}$$
