Powers problem

Calculus Level 3

A = 1 4 + 2 4 + 3 4 + 4 4 . . . B = 1 4 + 3 4 + 5 4 + 7 4 . . . A=1^{-4}+2^{-4}+3^{-4}+4^{-4}... \\ B=1^{-4}+3^{-4}+5^{-4}+7^{-4}...

Find A B \dfrac AB .

The answer is of the form p q \dfrac pq , where p p and q q are coprime positive integers. Give answer as p . q p.q . For example, if your answer is 99 100 \dfrac {99}{100} give it as 99.100.


The answer is 16.15.

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2 solutions

Chew-Seong Cheong
Feb 19, 2018

B = 1 4 + 3 4 + 5 4 + 7 4 + . . . = 1 4 + 2 4 + 3 4 + 4 4 + 5 4 + 6 4 + 7 4 + . . . ( 2 4 + 4 4 + 6 4 + 8 4 + . . . ) = 1 4 + 2 4 + 3 4 + 4 4 + . . . 2 4 ( 1 4 + 2 4 + 3 4 + 4 4 + . . . ) = A 1 16 A = 15 16 \begin{aligned} B & = 1^{-4} + 3^{-4} + 5^{-4} + 7^{-4} + ... \\ & = 1^{-4} + {\color{#3D99F6} 2^{-4}} + 3^{-4} + {\color{#3D99F6} 4^{-4}} + 5^{-4} + {\color{#3D99F6} 6^{-4}} + 7^{-4} + ... \color{#D61F06} - \left(2^{-4} + 4^{-4} + 6^{-4} + 8^{-4} + ...\right) \\ & = 1^{-4} + 2^{-4} + 3^{-4} + 4^{-4} + ... - 2^{-4} \left(1^{-4} + 2^{-4} + 3^{-4} + 4^{-4} + ...\right) \\ & = A - \frac 1{16} A = \frac {15}{16} \end{aligned}

A B = 16 15 p . q = 16.15 \implies \dfrac AB = \dfrac {16}{15} \implies p.q = \boxed{16.15} .

A = 1 4 + 2 4 + 3 4 + 4 4 . . . . . . . A=1^{-4}+2^{-4}+3^{-4}+4^{-4}....... B = 1 4 + 3 4 + 5 4 + 7 4 . . . . . . . B=1^{-4}+3^{-4}+5^{-4}+7^{-4}....... Therefore, A = B + 2 4 + 4 4 + 6 4 + 8 4 . . . . . . . A=B+2^{-4}+4^{-4}+6^{-4}+8^{-4}....... A = B + 2 4 ( 1 4 + 2 4 + 3 4 + 4 4 . . . . . . . ) A=B+2^{-4}(1^{-4}+2^{-4}+3^{-4}+4^{-4}.......) A = B + 2 4 ( A ) A=B+2^{-4}(A) A 2 4 ( A ) = B A-2^{-4}(A)=B A ( 1 2 4 ) = B A(1 -2^{-4})=B A ( 1 1 / 16 ) = B A(1-1/16)=B A ( 15 / 16 ) = B A(15/16)=B A / B = 16 / 15 A/B=16/15

=16.15 as required

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