Guess and Prove by induction a formula for the $n$-th element in a sequence $b_n$

Question

So I've been given a sequence.

The sequence $b_0,b_1,b_2$, ... is defined as follows:

$b_0 = 0$, $b_1 = 1/2$, and for integers $n \ge 2$, $b_n = \sqrt{b_{n-1}b_{n-2}} + \frac{3n}{2} - 1.$

My calculations for the first 5 values of n: $b_0 = 0, b_1 = 0.5, b_2 = 2, b_3 = 4.5, b_4 = 8, b_5 = 12.5$

from these values I've seen a pattern for any $n \ge 0$ and have been asked to guess a formula to produce them.

Namely that $n^2/2$ will give me all those values above and more,

Additionally I'm asked for Proof by induction that the guess (in my case $n^2/2$) is true for all $n \ge 0$

Below is some of my scratch work attempting to start the proof, not part of the question simply trying to show my thinking a bit:

So far I've got $b_{k+1} = \frac{(k+1)^2}{2}$

so $b_k + b_1$ would be $\frac{k^2}{2} + \frac{1}{2}$

Lost on where to go next.

Answer 1

I guess you are trying to prove that

if $b_0=0$, $b_1=1/2$ and $b_n=\sqrt{b_{n-1}b_{n-2}}+\frac{3n}{2}-1$, for $n\ge2$, then $b_n=n^2/2$, for all $n\ge0$.

This is true for $n=0$ and $n=1$. Suppose it is for all $m<n$, where $n\ge2$. Then $$ b_n=\sqrt{\frac{(n-1)^2}{2}\frac{(n-2)^2}{2}}+\frac{3n}{2}-1 $$ Can you finish?