Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

First of all I am a programmer, not a mathematician, so I may articulate what I am trying to say very poorly. I was working with powers of $2$ when I noticed a relationship I had never noticed before. Basically, if you look at the powers of $2$, every other power of two can be added together make a multiple of 10. Furthermore, if the two numbers are $2^{10}$ and $2^8$, adding them together will result in $2^7\cdot 10$. Put another way, $2^n + 2^{n-2} = 2^{n-3}\cdot 10$.

Does this pattern have a name? I am attaching an image just to clarify my poor explanation. Thanks all! $$ \begin{matrix} & & \vdash & 20 & \dashv & \vdash & 160 & \dashv & \vdash & 1280 & \dashv \\ 1 & 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 & 512 & 1024 \\ & \vdash & 10 & \dashv & \vdash & 80 & \dashv & \vdash & 640 & \dashv \\ \end{matrix} $$

share|improve this question
4  
I doubt it has a name because it's a fairly simple consequence of the definitions: $2^n+2^{n-2}=2^{n-3}(2^3+2^1)=2^{n-3}\times 10$. It's still a pretty cool pattern though! (And actually demonstrating that nobody has given it a name, if that is in fact the case, is likely to be quite difficult!) –  Matt Pressland Apr 10 '12 at 15:48
    
This picture needs a LaTeX-wiz. \rlap, \underbrace and \phantom... and all that. –  user2468 Apr 10 '12 at 16:02
    
I TeXified the picture. Please make sure I did not change it. –  user2468 Apr 10 '12 at 16:11
    
It seems to be the same! –  Brendon Dugan Apr 10 '12 at 16:17

1 Answer 1

Notice the following:

$$2^n+2^{n+2}=2^n(1+2^2)=2^n\cdot 5 = 10\cdot 2^{n-1}$$

For $n=8$ we have that indeed $2^n+2^{n+2}=2^8+2^{10}=2^7\cdot10$ as you have mentioned.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.