Is it possible to show that $a\cdot a^{\lfloor\frac{b}{2}\rfloor}\cdot a^{\lfloor\frac{b}{2}\rfloor} = a^b$ when $b$ is odd

Question

I have $a$ and $b$ and $b$ is odd $a$ is an integer and $b$ is a strictly positive integer. Is there a way I can show:

$a\cdot a^{\lfloor\frac{b}{2}\rfloor}\cdot a^{\lfloor\frac{b}{2}\rfloor} = a^b$

I know I can simplify to:

$a^{2\lfloor\frac{b}{2}\rfloor + 1}$ and that $2\lfloor\frac{b}{2}\rfloor + 1$ is in the form of an odd number $2k+1$. Is there any way I can simplify more?

Answer 1

Hint:

If $b$ is odd, $\lfloor \frac{b}{2}\rfloor = \frac{b-1}2.$ Substitute that and you can simplify the expression.