Recurrence Relation Puzzle 1
   

   

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Recurrence Relation Puzzle 1

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The Fibonacci Numbers are defined as F1 = F2 = 1 and Fn+1 = Fn + Fn-1.

Calculate the following infinite sum:

 F1      3F2      32F3          3nFn+1 
---  +  ----  +  ----- + ... + ----- + ...
 1        5        52           5n
Answer: 25

Solution:

Let A = F1 + (3/5)F2 + (32/52)F3 + ...

A = F1 + (3/5)F2 + (32/52)(F1+F2) + (33/53)(F2+F3) + ...

A = 1 + 3/5 + (32/52)F1 + (32/52)F2 + (33/53)F2 + (33/53)F3 + ...

A = 1 + (3/5)F1 + (32/52)F1 + (32/52)F2 + (33/53)F2 + (33/53)F3 + ...

Split A into two pieces, taking alternate terms for A1 and A2,
with A1 + A2 = A

A1 = 1 + (32/52)F1 + (33/53)F2 + ...
A2 = (3/5)F1 + (32/52)F2 + (33/53)F3 + ...

You can see that

A1 = 1 + (9/25)A, and
A2 = (3/5)A, and
A1+A2=A, so
A1 = (2/5)A = (10/25)A, so from the two equations for A1,
(10/25)A = 1 + (9/25)A

So A = 25

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