Compute $\sum_{i=0}^{2n} (-3)^i$ by splitting the series into two parts.

Question

Compute $\sum_{i=0}^{2n} (-3)^i$ by splitting the series into two parts.

How do I split it into two parts? All I can tell so far is that the sum is going to be a positive number (probably) because any $i * 2$ is even, and any negative number to an even power is positive.

Edit:

I split it into $\sum_{i=0}^{n} (-3)^{2i} + \sum_{i=0}^{n - 1} (-3)^{2i + 1}$ but I have no idea where t0 go from there.

Answer 1

Like you said, split the series into even and odd terms that is $$\begin{align} \sum_{i=0}^{2n} (-3)^i &=\sum_{i=0}^{n} (-3)^{2i}+\sum_{i=0}^{n-1} (-3)^{2i+1} \\ &=\sum_{i=0}^{n} 3^{2i}-\sum_{i=0}^{n-1} 3^{2i+1}\\ &= \sum_{i=0}^{n} 3^{2i}-\sum_{i=0}^{n} 3^{2i-1}\end{align}.$$

It is just a calculation now.

Answer 2

Hint If you're not sure about a series, it may be helpful to expand out a few of the beginning terms:

$$-3 + 9 - 27 + 81 - 243 + 729 \mp \dots - 3^{2n-1} + 3^{2n}$$

$$(-3 + 9) + (-27 + 81) + (-243 + 729) \mp \dots + (-3^{2n-1} + 3^{2n})$$

$$6 + 54 + 486 + \dots + 6 * 3^{2n-2}$$

$$6(1 + 9 + 81 + \dots + 9^{n-1})$$

--> Use geometric series formula

(Note that $3^{2n-2} = 9^{n-1}$)