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.

Hana Wehbi by Hana Wehbi , 51, USA

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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}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

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

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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

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