A size n determinant has all 1s on the main diagonal and the first superdiagonal. It has all 1s on the first subdiagonal. It has zeros in all other places.
Show that this determinant evaluates to
Fibonacci number F(n+1).
The determinant corresponding to n=5 is depicted below:
 1 1 0 0 0 
1 1 1 0 0 
 0 1 1 1 0  = 8 = F(6)
 0 0 1 1 1 
 0 0 0 1 1 
A graphic solution is easy as you see from the sample that
n=1*a 1*b+0*c0*d+...+0w (n size of the matrix and choosing the first row to calculate)
But a=n1
So: n=n1b
But b=(1)n2 (the other terms of the first column of this matrix are zeroes)
So: n=n1+n2
Then:
1=1=F(2)
2= 1 1
1 1 = 2= F(3)
So: 3=1+2 = F(2)+F(3) = F(4) and with recursion:
n = n1+n2 = F(n)+F(n1)= F(n+1)
Edited on April 19, 2016, 12:57 pm

Posted by armando
on 20160419 11:23:17 