A problem by krishna mandalika

This is how you do the grouping.

$(1+2-3)+(4+5-6)+(7+8-9)+...+(1999+2000-2001)$

$= 0+3+6+...+1998$

$= 3(1+2+...+666)$

$= 3 \times \displaystyle \sum_{n=1}^{666} n$

$= 3 \times \frac {(666)(667)}{2}$

$= 3 \times 333 \times 667 = \boxed {666333}$

Nice problem! @Krishna

By grouping..........
(1+2+-3)+(4+5-6)+...................+(1999+2000-2001)
=0 + 3 + 6 + ..................+1998
Sn=n(a+l)/2
n =1998/3 = 666,
Sn= 666(3+1998)/2 =666333(Ans.)