Double nth-step Fibonacci summation

log 8192 ( n = 1 1729 ( k = 1 F k n 2 k ) + 1 ) \large \log_{8192} \left( \sum_{n=1}^{1729} \left ( \sum_{k=1}^\infty \frac{F_k ^n}{2^k} \right) +1 \right)

let F k n F^n_k be the k th k^\text{th} term of the n n -step Fibonacci number. then evaluate the expression above.

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The answer is 133.

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1 solution

Jake Lai
Jul 6, 2015

Kind of a needless extension of my problem. It can be proven that k = 1 F k ( n ) 2 k = 2 n \displaystyle \sum_{k=1}^{\infty} \frac{F^{(n)}_{k}}{2^{k}} = 2^{n} .

Got a proof?

Pi Han Goh - 5 years, 9 months ago

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Simple rearrangement, there's a problem of mine with a proof.

Jake Lai - 5 years, 9 months ago

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link pls .......

Pi Han Goh - 5 years, 9 months ago

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