Guess and prove Formula for Sequence $b_n$ = $\sqrt{b_{n-1}b_{n-2}} + \frac{3n}{2} - 1.$

Question

Having a really hard time wrapping my head around how to approach this. written out the formula up to $b_5$ but not seeing a pattern at all, I've scoured my notes, google and so on. missing some critical piece of knowledge I can't seem find.

Appreciate any help.

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

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

Guess the formula for $b_n$ for all integers $n \ge 0.$

Prove By Induction that the guess is correct.

Answer 1

Hint: Compute the first few terms: $$ \begin{array}{rc} n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ b_n & 0 & 0.5 & 2 & 4.5 & 8 & 12.5 & 18 & 24.5 & 32 & 40.5 & 50 \\ 2b_n & 0 & 1 & 4 & 9 & 16 & 25 & 36 & 49 & 64 & 81 & 100 \\ \end{array} $$

Answer 2

Some of the terms:

$$\begin{align} b_0 &= 0\\ b_1 &= 1/2\\ b_2 &= 2\\ b_3 &= 9/2 = 4.5\\ b_4 &= 8\\ b_5 &= 25/2 = 12.5\\ b_6 &= 18 \end{align}$$

The pattern I'm noticing:

$$\begin{align} b_3 - b_2 &= 2.5\\ b_4 - b_3 &= 3.5\\ b_5 - b_4 &= 4.5\\ b_6 - b_5 &= 5.5 \end{align}$$

The jumps in the changes of is equal to $1$, i.e.$5.5 - 4.5 = 4.5 - 3.5 = ... = 1$.

This would in turn seem to suggest

$$b_n \propto \frac{n^2}{2}$$

since $(n^2/2)'' = 1$. We see $n^2$ because what is sort of a "discrete second derivative" of sorts (the changes in the changes) is nonzero, but the changes in the changes of the changes is $0$.

With that in mind, testing a few values seems to make it clear that any other potential terms (we could have a linear and constant term like any quadratic) are zero, so we might be able to safely assume

$$b_n = \frac{n^2}{2}$$

for all integers $n \geq 0$.

(Note: I'm not sure how to make the derivation rigorous. I'm just playing with some techniques I've used before to find relations in these sorts of situations, and haven't quite reached the point of discussing finding explicit formulas for recurrence relations in my classes.)