How to prove $a_1 = 2$, $a_2 = 4$ and $a_{n+1} = \frac{1}{3}(2a_n+a_{n-1})$ for all $n \geq2$

Question

Let $a_1 = 3$, $a_2 = 4$ and $a_{n+1} = \frac{1}{3}(2a_n+a_{n-1})$ for all $n \geq2$ Prove that for all positive integers $n$, $3 \leq a_n \leq4$

This was a practice problem in my textbook in the chapter of strong induction. It's not worth marks or anything I've just spent a while trying to figure how to prove this and I'm drawing a blank.

Can someone help?

Answer 1

Because by the assumption of the induction $$a_{n+1`}\geq\frac{1}{3}(2\cdot3+3)=3$$ and $$a_{n+1`}\leq\frac{1}{3}(2\cdot4+4)=4.$$ Also, check $n=2$ and $n=3$.

Answer 2

Hint

Do you know how to parameterise the points of a closed interval $[a,b]$?

Answer 3

$x^2-\frac{2}{3}x-\frac{1}{3}=(x-1)(x+\frac{1}{3})=0$ then $a_n=A(1)^n+B(\frac{1}{3})^n$.That $a_n=5-9(\frac{1}{3})^n$