The problem I have is to Prove that $F_{2n-1} + F_{2n+1} = F^2_{2n}+1$
So,this is of course the Fibonacci sequence 1, 2, 3, 5, 8, etc. This can be rewritten as $F_k = F_{k-1} + F_{k-2}$
I've passed the 1st step (small case) My induction hypothesis is that Assume $F_{2i-1} + F_{2i+1} = F^2_{2i}+1, 0 \le i \le k$
Now, I should show that $F_{2(k+1)-1} + F_{2(k+1)+1} = F^2_{2(k+1)}+1$ I don't really get how to do this though? This expands to $F_{2k+1} + F_{2k+3} = F^2_{2k+2}+1$
Should I change $F_{2k+1}$ to $F_{2k} + F_{2k-1}$? I tried playing around with this for an hour but it didn't come to anything. I know there's a simple solution but I'm missing it. Could someone help ? Thanks!
The relation $F_{2n-1}+F_{2n+1}=F_{2n}^2+1$ is incorrect and because of this you can not prove it!. Consider the following matrix that is well-know to $Q_p$ matrix $$ Q_2=\left[ \begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array} \right] \, . $$ with the induction on $n$, you can prove that the $n$th power of matrix $Q_2$, is in the following form $$ Q_2^n= \left[ \begin{array}{cc} F_{n-1} & F_{n} \\ F_{n} & F_{n+1} \end{array} \right] \Longrightarrow Q_2^{2n}= \left[ \begin{array}{cc} F_{2n-1} & F_{2n} \\ F_{2n} & F_{2n+1} \end{array} \right]\, . $$ In addition, the determinant of matrix $Q_2$ is $(-1)$ and because of this we conclude that $$ det(Q_2^{2n})={(-1)}^{2n}=1\, . $$ the above relation, results that $$ det( \left[ \begin{array}{cc} F_{2n-1} & F_{2n} \\ F_{2n} & F_{2n+1} \end{array} \right] )=1 \Longrightarrow F_{2n-1}F_{2n+1}-F_{2n}^2=1 $$ So, we have $$ F_{2n-1}F_{2n+1}=F_{2n}^2+1 $$ If you would like to know more about Fibonacci numbers, you can study the outstanding book "Fibonacci and Lucas Numbers with Applications" by Koshy that you can find lots of nice relations about Fibonacci numbers in this book.