$V_n = 3^{n}$
I need to calculate its summation where
$S = (V_0)^{2} + (V_1)^{2} + ... + (V_{n-1})^2 $
Obviously, you can just put a new iteration $(W_n)$ where $W_n = (V_n)^{2}$ You find it $W_n = 9^{n}$ and then in the end you find
$S = (9^{n} - 1) / 8$
But when I tried putting it like this
$S = 3^{2(0)} + 3^{2(1)} + ... + 3^{2(n-1)}$
I find $S = (3^{2n-1} - 1) / 2$
And if you put $n = 1$ or $n = 0$ in both summations we found, you clearly don't find the same value which means the second one is probably wrong but I don't understand why. It should be the same.
No, $(3^{2n-1}-1)/2$ would be $3^0+3^1+\cdots+3^{2n-2}$, whereas $S$ lacks the odd-exponent terms. As @zwim notes, the even terms are $S$ and the odd ones are $3S$, and their sum is $4S$. Note $4=(3^2-1)/(3-1)$.