MIT Mathematics Tournament (III)

Algebra Level 3

Find the sum of: 2 1 4 1 1 + 2 2 4 2 1 + 2 4 4 4 1 + 2 8 4 8 1 + \frac{2^1}{4^1-1}+\frac{2^2}{4^2-1}+\frac{2^4}{4^4-1}+\frac{2^8}{4^8-1}+\dots


The answer is 1.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Brian Moehring
Aug 7, 2018

The general term of the sum, indexed by n = 0 , 1 , 2 , n=0,1,2,\ldots is 2 2 n 2 2 2 n 1 = 2 2 n 1 2 2 2 n = k = 0 ( 1 2 ) ( 1 + 2 k ) 2 n \frac{2^{2^n}}{2^{2\cdot 2^n}-1} = \frac{2^{-2^n}}{1 - 2^{-2\cdot 2^n}} = \sum_{k=0}^\infty \left(\frac{1}{2}\right)^{(1+2k)2^n} so the sum in the problem is n = 0 k = 0 ( 1 2 ) ( 1 + 2 k ) 2 n = m = 1 ( 1 2 ) m \sum_{n=0}^\infty \sum_{k=0}^\infty \left(\frac{1}{2}\right)^{(1+2k)2^n} = \sum_{m=1}^\infty \left(\frac{1}{2}\right)^m where the equality is by rearranging (all the terms are positive, so that's no problem), using the fact that every positive integer is written uniquely as the product of a non-negative power of 2 2 and a positive odd number.

Therefore the sum is m = 1 ( 1 2 ) m = 1 2 1 1 2 = 1 \sum_{m=1}^\infty \left(\frac{1}{2}\right)^m = \frac{\frac{1}{2}}{1-\frac{1}{2}} = \boxed{1}

Is it possible to have 2 variables in one sum? k = 0 ( 1 2 ) ( 1 + 2 k ) 2 n \displaystyle \sum_{k=0}^\infty (\frac{1}{2})^{(1+2k)2^n} You have 2 variables; k and n

I got wrong for this question but I managed to write up to this (after that I was lost)--[:(]

n = 2 x ( 2 n 4 n 1 ) \sum_{n=2^x}^\infty (\dfrac{2^n}{4^n - 1})

:))

Syed Hamza Khalid - 2 years, 8 months ago

Log in to reply

It's not a sum over two variables: k k is the only bound variable in that sum. If n = 3 , n=3, the sum is k = 0 ( 1 2 ) ( 1 + 2 k ) 2 3 \sum_{k=0}^\infty \left(\frac{1}{2}\right)^{(1+2k)2^3} or if n = 12 , n=12, the sum is k = 0 ( 1 2 ) ( 1 + 2 k ) 2 12 . \sum_{k=0}^\infty \left(\frac{1}{2}\right)^{(1+2k)2^{12}}. This is what I was saying at the beginning with the first line simply being the n n th term in the original sum.

Brian Moehring - 2 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...