Question

Find the sum of the first four terms of the geometric sequence shown below. 4​, 4/ 3​, 4/9​, ...

Answer 1

It's evident that the first four terms are 4, 4/3, 4/9, and 4/27. So the fourth partial sum of the series is

$S_4=4+\dfrac43+\dfrac49+\dfrac4{27}$

It's as easy as adding up the fractions, but I bet this is supposed to be an exercise in taking advantage of the fact that the series is geometric and use the well-known formula for computing such a sum.

Multiply the sum by 1/3 and you have

$\dfrac13S_4=\dfrac43+\dfrac49+\dfrac4{27}+\dfrac4{81}$

Now subtracting this from $S_4$ gives

$S_4-\dfrac13S_4=4-\dfrac4{81}$

That is, all the matching terms will cancel. Now solving for $S_4$, you
have

$\dfrac23S_4=4\left(1-\dfrac1{81}\right)$
$S_4=6\left(1-\dfrac1{81}\right)$
$S_4=\dfrac{480}{81}=\dfrac{160}{27}$