A simple and quick question. Have been sitting over it for a while now but i can't get it right: Could someone just clarify how this transformation has been done?
$$\frac{(v+1)^2 2^{v^2}}{2^{v^2+2v+1}}=\frac{1}{2}(\frac{v+1}{2^v})^2$$
Well i understand that he canceled out the $2^{v^2}$ and then pulled the exponent out of the brackets, but where does that $\frac{1}{2} $ come from and where did the $2v+1$ in the exponent go.. any tips will be useful, thank you.
First, as you said, he (I'm assuming the person who posed the problem) cancelled $2^{v^2}$ from the numerator and denominator. This leaves you with $\frac{{(v+1)}^2}{2^{2v+1}}$. The $\frac 1 2$ comes from the denominator by taking out $\frac{1}{2^1}$, leaving $\frac{{(v+1)}^2}{2^{2v}}$. Simplifying $\frac{1}{2^{2v}}$ to $\frac{1}{(2^v)^2}$ yields the final result, $\frac 1 2\left(\frac{v+1}{2^v}\right)^2$.