Geometric series >$3+{{3}\over{-4}}+...+{{3}\over {(-4)^{n-1}}}+...$

Question

Find $S_1, S_2, S_3$ and $S_n$, and the sum of series if it converges.

$3+{{3}\over{-4}}+...+{{3}\over {(-4)^{n-1}}}+...$

My attempt:

$S_1=3$, $S_2={{9}\over {4}}$, $S_3={{33}\over {16}}$

The series converges, since it is geometric with $|r|={{1}\over {4}}<1$ and a=3, the sum is ${{12}\over{5}}$

But what about $S_n$ how can I find it?

Thanks

Answer 1

In general:

$1 + r + r^2 + ... + r^n = \dfrac{1 - r^{n-1}}{1-r}$

Applying that to your case, we have:

$S_n = 3\dfrac{1 - \left(\dfrac{1}{4}\right)^{n-1}}{\dfrac{5}{4}} = 12 \cdot \dfrac{1 - \left(\dfrac{1}{4}\right)^{n-1}}{5}$

Answer 2

HINT

We have that

$$S_n=\sum_0^n \frac{3}{(-4)^k}=3\sum_0^n \left(-\frac{1}{4}\right)^k$$

then refer to the geometric series sum.