Question

The sum of the series $1 - 2 + 3 - 4 + 5 + ...$ up to $1001$ terms is

• A. $500$
• B. $ - 500$
• C. $501$
• D. $ - 501$

Answer 1

Answer: (C)

Let $S = \left(1 - 2\right) + \left(3 - 4\right) + \left(5 - 6\right) + ... + \left(999 - 1,000 \right) +1001$
$= - \underbrace{(1 + 1 +... + 1)}_{500 \, times} + 1001 = 1001 - 500 = 501$
Alternative Solution :
$S = \left(1 + 3 + 5 + ... + 1001\right) - \left(2 + 4 + 6 + ... + 1000\right)$
= (first $501$ odd natural numbers) - $2$(sum of first $500$ natural numbers)
$= \left(501\right)^{2} - 500 \times 501$
$= \left(501\right) \left[501 - 500\right] = 501$
Short Cut Method :
$1 - 2 + 3 - 4 +...= - \frac{N}{2}$ if $N$ is even
$ = \frac{N +1}{2}$ if $N$ is odd
$\therefore $ Required sum $= \frac{1001 +1}{2} = \frac{1002}{2} = 501$