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 $S=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(999-1,000\right)+1001$
$=-\underset{500times}{\underbrace{(1+1+...+1)}}+1001=1001-500=501$
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 $=\frac{1001+1}{2}=\frac{1002}{2}=501$