Question

The expression
$(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2008})(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2007})-(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2008})(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2007})$ simplifies to

• A. 0
• B. $\frac{1}{2007}$
• C. $\frac{1}{2008}$
• D. $\frac{2}{2007}$

Answer 1

The correct option is C $\frac{1}{2008}$

$(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2008})(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2007})-(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2008})(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2007})$
Let $(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2008})=x$
and $(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2007})=y$
Then, $x(1+y)-y(1+x)$
= $x+xy-y-xy$
= $x-y$
= $(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2008})-(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2007})$
= $\frac{1}{2008}$