Answer
Geometric
Sum: $\dfrac{\sqrt 3(3^{25}-1)}{\sqrt 3-1}$
Work Step by Step
We are given the sequence:
$\{3^{n/2}\}$
Compute the ratio between two consecutive terms:
$\dfrac{a_{k+1}}{a_k}=\dfrac{3^{(n+1)/2}}{3^{n/2}}=3^{1/2}=\sqrt 3$
As the ratio between any consecutive terms is constant, the sequence is geometric.
Its elements are:
$a_1=3^{1/2}=\sqrt 3$
$r=\sqrt 3$
We determine the sum of the first 50 terms:
$S_n=a_1\cdot\dfrac{1-r^n}{1-r}$
$S_{50}=\sqrt 3\cdot\dfrac{1-\left(\sqrt 3\right)^{50}}{1-\sqrt 3}=\dfrac{\sqrt 3(3^{25}-1)}{\sqrt 3-1}$
