The Fibonacci numbers are defined as F₀ = F₁ = 1 and F_n+1 = F_n + F_n-1.

Calculate the following infinite sum:

 F0      3F1      32F2          3nFn 
---  +  ----  +  ----- + ... + ----- + ...
 1        5        52           5n

Solution:

Let A = F₀ + (3/5)F₁ + (3²/5²)F₂ + ...

A = F₀ + (3/5)F₁ + (3²/5²)(F₀+F₁) + (3³/5³)(F₁+F₂) + ...

A = 1 + 3/5 + (3²/5²)F₀ + (3²/5²)F₁ + (3³/5³)F₁ + (3³/5³)F₂ + ...

Split A into two pieces, with A₁ + A₂ = A

A₁ = 1 + (3²/5²)F₀ + (3³/5³)F₁ + ...
A₂ = 3/5 + (3²/5²)F₁ + (3³/5³)F₂ + ...

You can see that

A₁ = 1 + (9/25)A

Also, note that because 3/5 = (3/5)F₀, it follows that

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