Welcome 2016! Part 14

Algebra Level 4

∑ n = 1 ∞ ∑ k = 1 n − 1 k 2 n + k \large \displaystyle \sum_{n=1}^{\infty} \displaystyle \sum_{k=1}^{n-1} \dfrac{k}{2^{n+k}}

If the value of the series above is equal to A B \dfrac AB , where A A and B B are coprime positive integers, find the value of B 2 − A 2 B^2-A^2 .


The answer is 65.

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Anish Harsha by Anish Harsha , 20, India

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2 solutions

Akshat Sharda
Dec 20, 2015

S = ∑ n = 1 ∞ ∑ k = 1 n − 1 k 2 n + k S = ( 1 2 3 ) + ( 1 2 4 + 2 2 5 ) + ( 1 2 5 + 2 2 6 + 3 2 7 ) + … S = 1 2 3 + 1 2 4 + 3 2 5 + 3 2 6 + 6 2 7 + 6 2 8 + 10 2 9 + 10 2 10 + 15 2 11 + … S 2 = 1 2 4 + 1 2 5 + 3 2 6 + 3 2 7 + 6 2 8 + 6 2 9 + 10 2 10 + 10 2 11 + 15 2 12 + … S − S 2 = S 2 = 1 2 3 + 2 2 5 + 3 2 7 + 4 2 9 + 5 2 11 + … S 8 = 1 2 5 + 2 2 7 + 3 2 9 + 4 2 11 + 5 2 13 + … S 2 − S 8 = 3 S 8 = 1 2 3 + 1 2 5 + 1 2 7 + 1 2 9 + 1 2 11 + … 3 S 8 = 1 2 3 1 − 1 2 2 = 1 6 S = 4 9 ⇒ 9 2 − 4 2 = 65 \begin{aligned} S & = \displaystyle \sum_{n=1}^{\infty} \displaystyle \sum_{k=1}^{n-1} \dfrac{k}{2^{n+k}} \\ S & = \left(\frac{1}{2^3}\right)+\left(\frac{1}{2^4}+\frac{2}{2^5}\right)+\left(\frac{1}{2^5}+\frac{2}{2^6}+\frac{3}{2^7}\right)+\ldots \\ S & = \frac{1}{2^3}+\frac{1}{2^4}+\frac{3}{2^5}+\frac{3}{2^6}+\frac{6}{2^7}+\frac{6}{2^8}+\frac{10}{2^9}+\frac{10}{2^{10}}+\frac{15}{2^{11}}+\ldots \\ \frac{S}{2} & = \frac{1}{2^4}+\frac{1}{2^5}+\frac{3}{2^6}+\frac{3}{2^7}+\frac{6}{2^8}+\frac{6}{2^9}+\frac{10}{2^{10}}+\frac{10}{2^{11}}+\frac{15}{2^{12}}+\ldots \\ S-\frac{S}{2}=\frac{S}{2} & = \frac{1}{2^3}+\frac{2}{2^5}+\frac{3}{2^7}+\frac{4}{2^9}+\frac{5}{2^{11}}+\ldots \\ \frac{S}{8} & = \frac{1}{2^5}+\frac{2}{2^7}+\frac{3}{2^9}+\frac{4}{2^{11}}+\frac{5}{2^{13}}+\ldots \\ \frac{S}{2}-\frac{S}{8}=\frac{3S}{8} & = \frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}+\frac{1}{2^{11}}+\ldots \\ \frac{3S}{8} & = \frac{\frac{1}{2^3}}{1-\frac{1}{2^2}}=\frac{1}{6} \\ S & =\frac{4}{9} \\ \Rightarrow 9^2-4^2 & =\boxed{65} \end{aligned}

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In your 5th line, how do you know that S − S / 2 S - S/2 cancel off nicely to form an Arithmetic-Geometric Progression ? Or another way to put it is: how do you know the general term of the numerators of all the fractions in your third line are all in the form of

a n = 1 2 ⌊ n + 1 2 ⌋ ( ⌊ n + 1 2 ⌋ + 1 ) ? a_n = \dfrac12 \left\lfloor \dfrac{n+1}2 \right \rfloor \left( \left\lfloor \dfrac{n+1}2 \right \rfloor + 1\right) ?

The right way to solve this by noting that this is an absolutely convergent series, so we are allowed to rearrange the terms:

1 2 1 ( 1 2 2 + 1 2 3 + 1 2 4 + ⋯ ) + 2 2 2 ( 1 2 3 + 1 2 4 + 1 2 5 + ⋯ ) + 3 2 3 ( 1 2 4 + 1 2 5 + 1 2 6 + ⋯ ) + ⋯ \begin{aligned} && \dfrac1{2^1} \left( \dfrac1{2^2} + \dfrac1{2^3} + \dfrac1{2^4} + \cdots \right) \\ && + \dfrac2{2^2} \left( \dfrac1{2^3} + \dfrac1{2^4} + \dfrac1{2^5} + \cdots \right) \\ && + \dfrac3{2^3} \left( \dfrac1{2^4} + \dfrac1{2^5} + \dfrac1{2^6} + \cdots \right) \\ && + \cdots \end{aligned}

Upon simplifying, we can see that this is a sum of an AGP with infinite terms with a = 1 , d = 1 , r = 1 / 4 a = 1, d = 1, r = 1/4 .

Pi Han Goh - 5 years, 5 months ago

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Those numerators exist in pairs in the form n ( n + 1 ) 2 \frac{n(n+1)}{2} (triangular numbers).

So S − S 2 S-\frac{S}{2} forms AGP.

Akshat Sharda - 5 years, 5 months ago

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You need to show that it's true. Just because they appear to the be true for the first few terms does not mean that the rest of the terms follow.

Pi Han Goh - 5 years, 5 months ago

∑ n = 1 ∞ ∑ k = 1 n − 1 k 2 n + k = ( 1 2 3 ) + ( 1 2 4 + 2 2 5 ) + ( 1 2 5 + 2 2 6 + 3 2 7 ) + ( 1 2 6 + 2 2 7 + 3 2 8 + 4 2 9 ) + . . . = ( 1 2 3 + 2 2 5 + 3 2 7 + 4 2 9 + . . . ) ( 1 + 1 2 + 1 2 2 + 1 2 3 + . . . ) = 1 2 3 ( 1 + 2 2 2 + 3 2 4 + 4 2 6 + . . . ) ( 1 + 1 2 + 1 2 2 + 1 2 3 + . . . ) Sum of an AGP Sum of a GP = 1 2 3 ( 1 1 − 1 4 + 1 4 ( 1 − 1 4 ) 2 ) ( 1 1 − 1 2 ) = 1 2 3 ( 4 3 + 4 9 ) ( 2 ) = 4 9 \begin{aligned} \sum_{n=1}^\infty \sum_{k=1}^{n-1} \frac{k}{2^{n+k}} & = \left( \frac{1}{2^3} \right) + \left( \frac{1}{2^4} + \frac{2}{2^5} \right) + \left( \frac{1}{2^5} + \frac{2}{2^6} + \frac{3}{2^7} \right) + \left( \frac{1}{2^6} + \frac{2}{2^7} + \frac{3}{2^8} + \frac{4}{2^9} \right) + ... \\ & = \left( \frac{1}{2^3} + \frac{2}{2^5} + \frac{3}{2^7} + \frac{4}{2^9} +... \right) \left(1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} +... \right) \\ & = \frac{1}{2^3} \color{#3D99F6}{\left( 1 + \frac{2}{2^2} + \frac{3}{2^4} + \frac{4}{2^6} +... \right)} \color{#D61F06} {\left(1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} +... \right)} \quad \quad \small \color{#3D99F6}{\text{Sum of an AGP}} \quad \color{#D61F06}{\text{Sum of a GP}} \\ & = \frac{1}{2^3} \color{#3D99F6}{\left( \frac{1}{1 - \frac{1}{4}} + \frac{\frac{1}{4}}{\left(1 - \frac{1}{4}\right)^2} \right)} \color{#D61F06} {\left( \frac{1}{1 - \frac{1}{2}} \right)} \\ & = \frac{1}{2^3} \color{#3D99F6}{\left( \frac{4}{3} + \frac{4}{9} \right)} \color{#D61F06} {\left( 2 \right)} \\ & = \frac{4}{9} \end{aligned}

⇒ B 2 − A 2 = 9 2 − 4 2 = 65 \Rightarrow B^2-A^2 = 9^2 - 4^2 = \boxed{65}

More about AGP -- Arithmetic-Geometric Progression .

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