Sum of a finite geometric series $$\frac{1}{4} V^\frac{1}{4} + \frac{1}{4} V^\frac{1}{2} + \cdots + \frac{1}{4} V^{n-\frac{1}{4}} + \frac{1}{4} V^n =$$
$$\frac{1}{4} V^\frac{1}{4} \bigg(1 + V^\frac{1}{4} + V^\frac{1}{2} + \cdots + V^{n-\frac{1}{4}}\bigg) =$$
$$\frac{1}{4} V^\frac{1}{4} \bigg(\frac{1-V^{\big(n-\frac{1}{4}\big)+\frac{1}{4}}}{1-V^\frac{1}{4}}\bigg)$$
I understand that this is a geometric series with ratio v^1/4. Now what I don't understand is the third and last step where they have summed the series to obtain the expression in brackets.
$$S_n \equiv \sum_{k=0}^n r^k = \frac{1-r^{n+1}}{1-r}$$
I am now using the above formula. I have to sum from 0 to $n-\frac{1}{4}$ but when I substitute $r' = V^\frac{1}{4}$ and $n' = n-\frac{1}{4}$ in the expression on the right-hand side of the formula above I obtain the correct answer for the denominator, but for the numerator I get $V^{\frac{1}{4} (n-\frac{1}{4} + 1)} = V^{\frac{n}{4} + \frac{3}{16}}$. I do not understand how one can obtain $V^(n-\frac{1}{4} + \frac{1}{4}$. Any help would be much appreciated.


We obtain
  \begin{align*}
\frac{1}{4}&v^{\frac{1}{4}}\left(1+v^{\frac{1}{4}}+v^{\frac{1}{2}}+\cdots+v^{n-\frac{1}{4}}\right)\\
&=\frac{1}{4}v^{\frac{1}{4}}\left(1+v^{\frac{1}{4}}+v^{\frac{2}{4}}+\cdots+v^{\frac{4n-1}{4}}\right)\\
&=\frac{1}{4}v^{\frac{1}{4}}\left(1+v^{\frac{1}{4}}+\left(v^{\frac{1}{4}}\right)^2+\cdots+\left(v^{\frac{1}{4}}\right)^{4n-1}\right)\\
&=\frac{1}{4}v^{\frac{1}{4}}\cdot\frac{1-\left(v^{\frac{1}{4}}\right)^{4n}}{1-v^{\frac{1}{4}}}\\
\\
&=\frac{1}{4}v^{\frac{1}{4}}\cdot\frac{1-v^n}{1-v^{\frac{1}{4}}}
\end{align*}

