Sums of Squares of Fibonacci Numbers

Theorem
If  
\[F_{n-1}, \; F_n, \; F_{n+1}, \; n ge 2\]
  are consectutive Fibonacci numbers then  
\[F^2_n=F_{n+1}F_n-F_nF_{n-1}\]
.
Proof
The Fibonacci sequemce is defined as  
\[F_1=F_2=1, \; F_{n+1}=F_n+F_{n-1}\]
.
Multiply the recurrence relation by  
\[F_n\]
  to get  
\[F_{n+1}F_n=F^2_n+F_{n-1}F_n\]
  and rearrange to get  
\[F^2_n=F_{n+1}F_n-F_nF_{n-1}\]
.
To see how this might be useful, consider  
\[F^2_1+F^2_2+F^2_3+...+F^2_{10}\]
.
\[\begin{equation} \begin{aligned} F^2_1+F^2_2+F^2_3+...+F^2_{10} &= 1^2+(F_3F_2-F_2F_1)+(F_4F_3-F_3F_2) \\ &+ (F_5F_4-F_4F_3)+...+(F_{10}F_9-F_9F_8) \\ &+ (F_{11}F_{10}-F_{10}F_9) \\ &= F_{11}F_{10}=89 \times 55=4895 \end{aligned} \end{equation}\]

Add comment

Security code
Refresh