I have the following exercise:
"Use strong induction to prove that $f_1^2 + f_2^2 + \cdots + f_n^2 = (f_n)(f_{n+1})$ where $f_n$ in the nth Fibonacci number."
This is what I have done:
Fibonacci sequence - $f_n = f_{n-1} + f_{n-2}, f_1=1, f_2=1$
$f_3=f_2+f_1=2; f_4=f_3+f_2=3; f_5=f_4+f_3=5$ and so on.
*** BASIS STEP:
for $n=1: f_1^2 = f_1f_2; 1=1$
for $n=2: f_1^2 + f_2^2 = f_2f_3; 2=2$
for $n=3: f_1^2 + f_2^2 + f_3^2 = f_3f_4; 6=6$
Therefore $P(n)$ is true for n=1,2,3.
*** INDUCTIVE STEP:
Please help how to set this step.
Now you need to show that $$f_1^2 + \cdots + f_{n+1}^2 = f_{n+1}f_{n+2}$$ is true if $$f_1^2 + \cdots + f_m^2 = f_m f_{m+1}$$ is true for all $m \leq n$.
We proceed by using the equality for $m=n$ and then applying the property $f_{n+2} = f_n + f_{n+1}$. \begin{align*} f_1^2 + \cdots + f_n^2 + f_{n+1}^2 & = f_{n}f_{n+1} + f_{n+1}^2 \\ & = f_{n+1}(f_n+f_{n+1}) \\ & = f_{n+1}f_{n+2} \end{align*}
And we're done. You should have noticed that we only required the truth of the equation for $m=n$, so a normal induction would have sufficed.