The first three determinants of a 1,3,1 matrix:

\[ S_1 = \begin{vmatrix} 3 \end{vmatrix} \]

\[ S_2 = \begin{vmatrix} 3 & 1 \\ 1 & 3 \end{vmatrix} \]

\[ S_3 = \begin{vmatrix} 3 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{vmatrix} \]

Show that \(S_n\) is the Fibonacci number \(F_{2n+2}\) by proving \(F_{2n+2} = 3F_{2n} - F_{2n-2}\). Keep using Fibonacci's rule \(F_k = F_{k-1} + F_{k-2}\) starting with \(k = 2n + 2\).

Attempt:

First, we can see via cofactors that

\[ \begin{aligned} S_1 &= 3 = F_4 \\ S_2 &= 9 - 1 = 8 = F_6 \\ S_3 &= 3S_2 - \begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} = 3S_2 - S_1 \end{aligned} \]

In general, we can see that

\[ \begin{aligned} S_n = 3S_{n-1} - S_{n-2} \end{aligned} \]

Next, setting \(k = 2n + 2\),

\[ \begin{aligned} F_{2n+2} &= F_{2n+1} + F_{2n} = (F_{2n} + F_{2n-1}) + F_{2n} \\ &= (3F_{2n} - F_{2n}) + F_{2n-1} \\ &= (3F_{2n} - F_{2n-1} - F_{2n-2}) + F_{2n-1} \\ \therefore F_{2n+2} &= 3F_{2n} - F_{2n-2} \\ \end{aligned} \]

Now, if \(S_n = F_{2n+2}\), then \(S_{n-1} = F_{2(n-1)+2} = F_{2n}\) and similarly \(S_{n-2} = F_{2n-2} \). But as demonstrated above this turns out to be the case for \(S_1 = F_4\) and \(S_2 = F_6\). Therefore, \(S_n\) is indeed the Fibonnaci number \(F_{2n+2}\). Interestingly, \(S_n\) is a sequence that produces every second Fibonnaci number !

# posted by rot13(Unafba Pune) @ 9:10 PM