Fibonacci 2021 Series

Algebra Level 4

∑ n = 0 ∞ 2 n F n ( 2021 + 4084445 ) n = 1 a \sum _{ n=0 }^{ \infty }{ \frac { { 2 }^{ n }{ F }_{ n } }{ { \left( 2021+\sqrt { 4084445 } \right) }^{ n } } } =\frac { 1 }{ a } F n { F }_{ n } is a Fibonacci Number. Find the value of a a


The answer is 2020.

Syed Shahabudeen by Syed Shahabudeen , 23, India

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1 solution

Syed Shahabudeen
Jun 1, 2020

∑ n = 0 ∞ 2 n F n ( 2021 + 4084445 ) n = ∑ n = 0 ∞ F n x n where x = 2021 + 4084445 2 = F 0 + F 1 x + ∑ n = 2 ∞ F n x n ∑ n = 2 ∞ F n x n = 1 x 2 ∑ n = 0 ∞ F n + 2 x n ; F n + 2 = F n + 1 + F n = F 0 + F 1 x + 1 x 2 ∑ n = 0 ∞ F n + 1 x n + 1 x 2 ∑ n = 0 ∞ F n x n = F 0 + F 1 x + 1 x ∑ n = 0 ∞ F n + 1 x n + 1 + 1 x 2 ∑ n = 0 ∞ F n x n ∑ n = 0 ∞ F n + 1 x n + 1 = ∑ n = 0 ∞ F n x n ; F 0 = 0 , F 1 = 1 = 1 x + 1 x ∑ n = 0 ∞ F n x n + 1 x 2 ∑ n = 0 ∞ F n x n ∑ n = 0 ∞ F n x n ( 1 − 1 x − 1 x 2 ) = 1 x ∑ n = 0 ∞ F n x n = x x 2 − x − 1 ∑ n = 0 ∞ F n ( 2021 + 4084445 2 ) n = ( 2021 + 4084445 2 ) ( 2021 + 4084445 2 ) 2 − ( 2021 + 4084445 2 ) − 1 = 1 2020 \begin{aligned} \sum _{ n=0 }^{ \infty }{ \frac { { 2 }^{ n }{ F }_{ n } }{ { \left( 2021+\sqrt { 4084445 } \right) }^{ n } } } &= \sum _{ n=0 }^{ \infty }{ \frac { { F }_{ n } }{ { x }^{ n } } } &&&&&&&&&&&&&&&\textcolor{#20A900}{\textrm{where}\quad{ x }=\frac { 2021+\sqrt { 4084445 } }{ 2 } } \\&={ F }_{ 0 }+\frac { { F }_{ 1 } }{ { x } } +\sum _{ n=2 }^{ \infty }{ \frac { { F }_{ n } }{ { x }^{ n } } } &&&&&&&&&&&&&&&\textcolor{#20A900} {\sum _{ n=2 }^{ \infty }{ \frac { { F }_{ n } }{ { x }^{ n } } } =\frac { 1 }{ { x }^{ 2 } } \sum _{ n=0 }^{ \infty }{ \frac { { F }_{ n+2 } }{ { x }^{ n } } } ;\quad { F }_{ n+2 }={ F }_{ n+1 }+{ F }_{ n } } \\&={ F }_{ 0 }+\frac { { F }_{ 1 } }{ x } +\frac { 1 }{ { x }^{ 2 } } \sum _{ n=0 }^{ \infty }{ \frac { { F }_{ n+1 } }{ { x }^{ n } } +\frac { 1 }{ { x }^{ 2 } } \sum _{ n=0 }^{ \infty }{ \frac { { F }_{ n } }{ { x }^{ n } } } } \\ &= { F }_{ 0 }+\frac { { F }_{ 1 } }{ x } +\frac { 1 }{ { x } } \sum _{ n=0 }^{ \infty }{ \frac { { F }_{ n+1 } }{ { x }^{ n+1 } } +\frac { 1 }{ { x }^{ 2 } } \sum _{ n=0 }^{ \infty }{ \frac { { F }_{ n } }{ { x }^{ n } } } } &&&&&&&&&&&&&&&\textcolor{#E81990}{\sum _{ n=0 }^{ \infty }{ \frac { { F }_{ n+1 } }{ { x }^{ n+1 } } =\sum _{ n=0 }^{ \infty }{ \frac { { F }_{ n } }{ { x }^{ n } } } }\quad;{ F }_{ 0 }=0 ,{ F }_{ 1 }=1 }\\&=\frac { 1 }{ x } +\frac { 1 }{ { x } } \sum _{ n=0 }^{ \infty }{ \frac { { F }_{ n } }{ { x }^{ n } } +\frac { 1 }{ { x }^{ 2 } } \sum _{ n=0 }^{ \infty }{ \frac { { F }_{ n } }{ { x }^{ n } } } }\\\sum _{ n=0 }^{ \infty }{ \frac { { F }_{ n } }{ { x }^{ n } } \left( 1-\frac { 1 }{ x } -\frac { 1 }{ { x }^{ 2 } } \right) } &=\frac { 1 }{ x }\\\sum _{ n=0 }^{ \infty }{ \frac { { F }_{ n } }{ { x }^{ n } } } &=\frac { x }{ { x }^{ 2 }-x-1 }\\\sum _{ n=0 }^{ \infty }{ \frac { { F }_{ n } }{ { \left( \frac { 2021+\sqrt { 4084445 } }{ 2 } \right) }^{ n } } } &=\frac { \left( \frac { 2021+\sqrt { 4084445 } }{ 2 } \right) }{ { \left( \frac { 2021+\sqrt { 4084445 } }{ 2 } \right) }^{ 2 }-\left( \frac { 2021+\sqrt { 4084445 } }{ 2 } \right) -1 }\\&=\frac { 1 }{ 2020 } \end{aligned} a = 2020 \boxed{a=2020}

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