<s>[INST] How do you solve the summation of $2-4+8-16+32- \dots 2^{48}$? This is a summation problem but I can't seem to figure out how to solve this with the mix of subtraction and addition.
 [/INST] $$S=2-4+8-16+32- \dots -2^{48} = (-1)(-2+4-8+16-32+ \dots +2^{48}) $$
$$ = (-1)\left((-2)^1+(-2)^2+(-2)^3+ \dots +(-2)^{48}\right)$$
which is a geometric series with first term $a=-2$ and common ratio $r=-2$ and $n=48$ terms. $S$ then becomes
$$S = (-1)\times \frac{a(1-r^{n+1})}{1-r} = -1\times \frac{-2(1-(-2)^{49})}{1-(-2) }= \frac{2(2^{49}+1)}{3}$$
Another way to look at this is to split the overall sum up into a positive part and a negative part:
$$S=2-4+8-16+...-2^{48} = (2+8+32+\dots +2^{47})-(4+16+64+\dots + 2^{48})$$ which are both geometric series with all positive terms
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