$1 - 2 + 3 -4 + \cdots - 98 + 99 =$
$\mathrm{\textbf{(A)} \ -50 } \qquad \mathrm{\textbf{(B)} \ -49 } \qquad \mathrm{\textbf{(C)} \ 0 } \qquad \mathrm{\textbf{(D)} \ 49 } \qquad \mathrm{\textbf{(E)} \ 50 }$
Solution 1
If we group consecutive terms together, we get $(-1) + (-1) + \cdots + 99$, and since there are 49 pairs of terms the answer is $-49 + 99 = 50 \Rightarrow \mathrm{\textbf{(E)}}$.

Solution 2
( Similar to Solution 1 )
If we rearranged the terms, we get $1+3-2+5-4 \cdots + 99-98$ then $1 + 1 + \cdots + 1$, and since there are 49 pairs of terms and the $1$ in the beginning the answer is $1+49 = 50 \Rightarrow \mathrm{\textbf(E)}$.

Solution 3
Let $1 - 2 + 3 -4 + \cdots - 98 + 99 = S$.
Therefore, $S=99-98+97-\cdots -4+3-2+1$
We add:
$2S=100-100+100-100\cdots +100=100$
$S=50\Rightarrow \mathrm{\textbf{(E)}}$


Solution 4
We proceed with addition, 1 -2 + 3 -4.... Once done we find $\mathrm{\textbf{(E)}}$