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.

Split it into a positive series and a negative series:
$\color\red{2}-\color\green{4}+\color\red{8}-\color\green{16}+\color\red{32}-\ldots-\color\green{2^{48}}=$
$\color\red{2^1}-\color\green{2^2}+\color\red{2^3}-\color\green{2^4}+\color\red{2^5}-\ldots-\color\green{2^{48}}=$
$\color\red{\sum\limits_{n=0}^{23}2^{2n+1}}-\color\green{\sum\limits_{n=0}^{23}2^{2n+2}}=$
$\color\red{2\sum\limits_{n=0}^{23}2^{2n}}-\color\green{4\sum\limits_{n=0}^{23}2^{2n}}=$
$\color\red{2\sum\limits_{n=0}^{23}4^n}-\color\green{4\sum\limits_{n=0}^{23}4^n}=$
$-2\sum\limits_{n=0}^{23}4^n=$
$-2\cdot\dfrac{4^{24}-1}{4-1}$
