Sum of 1+2-3-4... - 99-100?

$S=1+(2-3-4+5)+(6-7-8+9)+\cdots +98-99-100$ $=1+0+0+\cdots +98-99-100$ $=(1+98-99)-100=0-100=-100$

$\begin{aligned} S & = \color{#3D99F6}1+2-3-4\color{#D61F06}+5+6-7-8\color{#3D99F6}+9+10-11-12\color{#D61F06}+ \cdots \color{#3D99F6}+93+94-95-96\color{#D61F06}+97+98-99-100 \\ & = \underbrace{\color{#3D99F6}-4\color{#D61F06}-4\color{#3D99F6}-4\color{#D61F06}+ \cdots \color{#3D99F6}-4\color{#D61F06}-4}_{25 \times (-4)} \\ & = \boxed{-100} \end{aligned}$

(1-3)+( 2-4 )+( 5-7 )+(6-8).... =50 × (-2)

If you observe carefully.. -2+4-6...=2-2=0

-3+9-15...=3-3=0 So we can assume all numbers between 1and 100 cancel each other...also if x is a number in the series (which somehow is not cancelled), then next number would be -(x+1) Sum of those 2 is x-x-1 = -1 We remain with -100 and 1 and -1 So again 1 and -1 cancel out leaving -100 as the Answer