Finding formula for the nth partial sum A few days ago, I asked for some clarification about pattern recognition and the n-th partial sum for infinite series.  Although the explanation given was top-notch (thanks again), I'm still having difficulty with the homework.  The one I'm asking about tonight is the following sequence: $1 - 2 + 4 - 8 + ... + (-1)^{n-1}2^{n-1} + ...$
In my efforts to solve this problem, this is what I've gotten thus far:
$$
\begin{array}{lcc}
\textrm{Parial Sum} & \textrm{Value} & \textrm{Suggested Expression} \\
s_1 = 1 & 1 & ?? \\
s_2 = 1 - 2 & -1 & ??\\
s_3 = 1 - 2 +4 & 3 & ??\\
s_4 = 1 -2 +4 -8 & -5 & ??\\
\end{array}
$$
As you can see from the question marks where suggested expressions might be, I'm struggling to find the pattern.  What I do know is the formula to compute $a_n$, but I haven't discerned the pattern for the n-th partial sum.  Because this is a power of 2 series, I see that the magnitude between the values of each sum is exactly the power of 2 for the next n-1.  That is, the distance between 1 and -1 is $2^1$ and the distance between -1 and 3 is $2^2$.  I think that within this is the key to figuring this out.  Nevertheless, the solution eludes me and I need a hint.
One of my attempts was $(-1)^{n-1}*2(\frac{1}{2^{n-1}})$ which worked for the first two partial sums but then fell apart miserably.  While typing this up, I just made the further discovery that starting with $s_2$ each partial sum is equal to $2^n - m$ where m is the same number twice.  I know that probably doesn't make sense but $s_2, s_3$ are both equal to $2^n - 5$ and $s_4, s_5$ are both $2^n - 21$.  The next two are $2^n - 85$.  That can't be coincidence.
Please, help me see what I'm missing or help me to understand how I should set this up to find the pattern for the n-th partial sum.
Thanks,
Andy
 A: Later data often show patterns better than early data, so extend your table of partial sums a bit:
$$\begin{array}{rcc}
n:&1&2&3&4&5&6&7&8&9\\
s_n:&1&-1&3&-5&11&-21&43&-85&171
\end{array}$$
Ignoring the signs, it appears that the numbers in the bottom line are approximately doubling each time. Moreover, still ignoring signs, adjacent partial sums add up to a power of $2$: $|s_1|+|s_2|=2^1$, $|s_2|+|s_3|=2^2$, $|s_3|+|s_4|=2^3$, and apparently in general $|s_n|+|s_{n+1}|=2^n$. (If you go back to the definition of the partial sums, you’ll see why this happens.)
If $|s_n|+|s_{n+1}|=2^n$ and $|s_{n+1}|\approx 2|s_n|$, then $3|s_n|\approx 2^n$; this suggests that we should compare $3|s_n|$ with $2^n$:
$$\begin{array}{rcc}
n:&1&2&3&4&5&6&7&8&9\\
s_n:&1&-1&3&-5&11&-21&43&-85&171\\
3|s_n|:&3&3&9&15&33&63&129&255&513\\
2^n:&2&4&8&16&32&64&128&256&512
\end{array}$$
That pattern’s pretty clear: apparently $3|s_n|=2^n+1$ if $n$ is odd, and $3|s_n|=2^n-1$ if $n$ is even. Those cases can be combined as $3|s_n|=2^n+(-1)^{n+1}$, since $(-1)^{n+1}$ is $1$ when $n$ is odd and $-1$ when $n$ is even. And the algebraic sign of $s_n$ appears to be that of $(-1)^{n+1}$, so if these patterns are real,
$$\begin{align*}
s_n&=\frac{(-1)^{n+1}}3\left(2^n+(-1)^{n+1}\right)\\
&=\frac{(-1)^{n+1}2^n}3+\frac{(-1)^{2n+2}}3\\
&=\frac{(-1)^{n+1}2^n+1}3\\
&=\frac{1-(-2)^n}3\;.
\end{align*}$$
This result can then be proved by mathematical induction, but I suspect that you’re not expected to go that far.
If you’ve already learned the summation formula for finite geometric series, you can apply it to get $s_n$ without looking at any patterns at all, and it’s something that you should learn as soon as possible if you don’t already know it. However, skill at pattern-recognition is useful anyway, so I thought that it might be useful to see how the problem can be attacked in that way as well.
A: We have
$$1+r+r^2 + \cdots + r^{n-1} = \dfrac{1-r^n}{1-r}, \,\,\,\,\, \forall r \neq 1 \tag{$\star$}$$
In your case, $r=-2$.

 Hence, we get that the partial sum is$$s_n = \dfrac{1-(-2)^n}3$$

You may be interested in this question for more details on why $(\star)$ is true and what happens to its infinite sum, when $\vert r \vert < 1$.
A: We know that the common ratio between terms is $-2$, we can see that from the series:
$$S = 1 + (-2)1 + (-2)(-2) + (-2)4 + (-2)(-8) + ...$$
This means that the formula for the sum is:
$$\begin{align*}
 S&= \frac{1 - (-2)^n}{1 - (-2)}\\ 
 &=  \frac{1 - (-2)^n}{3}
\end{align*}$$
A: As we know, the $n^{th}$ partial sum for a given series is the sum of the first  $n$ terms of the series. In our case the $n^{th}$ partial sum becomes 
$$S_{n}=1-2+2^2-2^3+\cdots+(-1)^{n-1} 2^{n-1}. $$
Multiplying the above equation by 2 we get
$$2S_{n}=2-2^2+2^3-2^4+\cdots+(-1)^{n-1} 2^{n}.$$
Adding the above two equations simultinously, we obtain 
$$S_{n}=\frac{1+(-1)^{n-1} 2^{n}}{3}.$$
After simplifing, the $n^{th}$ partial sum  formula interms of $n$ becomes
$$S_{n}=\frac{1- (-2)^{n}}{3} .$$
