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