🧩 Concept

Consider \(N=\dfrac{1}{3}+\dfrac{1}{3^{2}}+\dfrac{2}{3^{3}}+\dfrac{3}{3^{4}}+\dfrac{5}{3^{5}}+\dots\) where numerators follow Fibonacci \(1,1,2,3,5,\dots\).

Goal: compute \(15N\) lightning-fast.

⚡ Shortcut (multiply & add)

1. Write \(N\) and \(3N\):
   \(N=\frac{1}{3}+\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\cdots\)
   \(3N=1+\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\cdots\)
2. Multiply again: \(9N=3+1+\frac{2}{3}+\frac{3}{3^2}+\cdots\)
3. After constants, each numerator matches the sum of the previous two ⇒ the tail of \(9N\) equals \((N+3N)\).

📐 Generator Check (1 line)

Since \( \sum_{n\ge1}F_n x^n=\dfrac{x}{1-x-x^2}\), take \(x=\tfrac13\): \(N=\dfrac{\tfrac13}{1-\tfrac13-\tfrac19}=\dfrac{3}{5}\) (confirms the shortcut).

🧪 Base-4 Variant

Evaluate \(S=\dfrac{1}{4^{2}}+\dfrac{1}{4^{3}}+\dfrac{2}{4^{4}}+\dfrac{3}{4^{5}}+\cdots\).

Let \(T=\sum_{n\ge1}\dfrac{F_n}{4^n}=\dfrac{\tfrac14}{1-\tfrac14-\tfrac1{16}}=\dfrac{4}{11}\). Then \(S=\dfrac14 T=\dfrac{1}{11}\).

📝 Quick Quiz

Let \(N'=\dfrac{1}{4}+\dfrac{1}{4^{2}}+\dfrac{2}{4^{3}}+\dfrac{3}{4^{4}}+\cdots\) (Fibonacci over powers of 4). What is \(11N'\)?

🎥 Video Walkthrough