How to find sum of this series using GP formula??? Given G.P:- $1 + 2^{1/n} + 2^{2/n} + . . . + 2^{(N-1)/N}$
My approach:-
We know, $S_{n} = \frac{a(r^n - 1)}{r - 1}$
Hence, here as we have n-1 terms, the formula will be:-
$S_{n-1} = \frac{a(r^{n-1} - 1)}{r - 1}$
We can see, $a = 1$, $r = 2^{1/n}$
Hence, $S_{n-1} = \frac{2^{(n-1)/n} - 1}{2^{1/n} - 1}$
Which is not same as the answer which is:- $\frac{1}{2^{1/n} - 1}$
Am I missing something?
Any help is appreciated.
Thank you!

Answer edited.
The sum of the first $N$ terms of a geometric series with first term $a$ and common ratio $r$ is: $a + ar + \ldots + ar^{N-1} =  \frac{a(1-r^N)}{1-r}.$
To see that this finite series has $N$ terms and not $N-1$ terms, write it as:
$$a + ar + \ldots + ar^{N-1} = ar^0 + ar^1 + \ldots + ar^{N-1} = ar^{-1}\underbrace{ \left( r^1 + r^2 + \ldots + r^N \right)}_{N \text{ terms}}\ . $$
Assuming you meant your sequence to be: $1 + 2^{1/N} + 2^{2/N} + . . . + 2^{(N-1)/N},$ then your sequence is the above with $a=1$ and $r = 2^{1/N}.$
