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+\frac43+\frac49+\frac4{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


`\frac13S_4=\frac43+\frac49+\frac4{27}+\frac4{81}`

Now subtracting this from
`S_4` gives


`S_4-\frac13S_4=4-\frac4{81}`

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


`\frac23S_4=4\left(1-\frac1{81}\right)`

`S_4=6\left(1-\frac1{81}\right)`

`S_4=(480)/(81)=(160)/(27)`