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.
 A: Hint: This is almost the geometric series $\displaystyle\sum_{k=1}^{48} (-2)^k$.
A: No need to use the formula for the geometric series.
$$\begin{align}
S&=2-4+8-16+32- \dots -2^{48}\\
2\,S&=4-8+16-32+\dots+2^{48}-2^{49}
\end{align}$$
Add them and voilà!
A: Hint:
Grouping the terms in pairs, this is
$$-2-8-32\cdots-2^{47}=-2(1+4+4^2\cdots+4^{23}).$$
A: Split it into a positive series and a negative series:
$\color\red{2}-\color\green{4}+\color\red{8}-\color\green{16}+\color\red{32}-\ldots-\color\green{2^{48}}=$
$\color\red{2^1}-\color\green{2^2}+\color\red{2^3}-\color\green{2^4}+\color\red{2^5}-\ldots-\color\green{2^{48}}=$
$\color\red{\sum\limits_{n=0}^{23}2^{2n+1}}-\color\green{\sum\limits_{n=0}^{23}2^{2n+2}}=$
$\color\red{2\sum\limits_{n=0}^{23}2^{2n}}-\color\green{4\sum\limits_{n=0}^{23}2^{2n}}=$
$\color\red{2\sum\limits_{n=0}^{23}4^n}-\color\green{4\sum\limits_{n=0}^{23}4^n}=$
$-2\sum\limits_{n=0}^{23}4^n=$
$-2\cdot\dfrac{4^{24}-1}{4-1}$
A: HINT:


*

*Start with $2^1+2^3+2^5+...+2^{47}$

*Then do $2^2+2^4+...+2^{48}$

*Now subtract both of them


You can accomplish 1. and 2. by using geometric progression sum formula as:
$ar^0+ar^1+ar^2+...+ar^{n-1}=a\dfrac{r^{n}-1}{r-1}$
More read : http://en.wikipedia.org/wiki/Geometric_progression
A: $$S=2-4+8-16+32- \dots -2^{48} = (-1)(-2+4-8+16-32+ \dots +2^{48}) $$
$$ = (-1)\left((-2)^1+(-2)^2+(-2)^3+ \dots +(-2)^{48}\right)$$
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
A: Compare with the non-alternating series:
$$S=2+4+8+16+32\dots+2^{48}.$$
You solve it by noting that
$$2S=4+8+16+32\dots+2^{48}+2^{49},$$
shares all terms with $S$, but the two extremes ones, and by subtraction,
$$2S-S=S=-2+2^{49}.$$

The process is identical for the alternating series:
$$S=2-4+8-16+32\dots-2^{48}.$$
You solve it by noting that
$$-2S=-4+8-16+32\dots-2^{48}+2^{49}.$$
shares all terms with $S$, but the two extremes ones, and by subtraction,
$$-2S-S=-3S=-2+2^{49}.$$
