Question

Evaluate the series 12 + 6 + 3 + . . . ?

Answer 1

`24`

Explanation:

We're going to want to find a general summation for this series.

At a glance, it does not look arithmetic, as the terms we're given don't each share a common difference. However, it looks like a geometric series. Let's find the common ratio `r` by dividing each given term by the previous term:

`6/12=1/2`

`3/6=1/2`

So, `r=1/2.`

Now, the general form for a geometric series is

`sum_(n=1)^ooa(r)^(n-1)` where `n` represents the `nth` term, `a` is the first term, and `r` is the common ratio.

From what we're given, `a=12,` and we found `r=1/2` so we get

`sum_(n=1)^oo12(1/2)^(n-1)`

Now, the infinite geometric series has the following value if `|r|<1:`

`sum_(n=1)^ooa(r)^(n-1)=a/(1-r)`. For our case, `|r|=1/2<1,` so we can find the value:

`sum_(n=1)^oo12(1/2)^(n-1)=12/(1-1/2)=12/(1/2)=12*2=24`