2002 AMC 10P Problems/Problem 23

Problem

Let

and

Find the integer closest to

Solution 1

Start by subtracting and and group those with a common denominator together, leaving and to the side.

$$\begin{array}{rl}a-b& =(\frac{1^2}{1}+\frac{2^2}{3}+\frac{3^2}{5}+\dots +\frac{{1001}^2}{2001})-(\frac{1^2}{3}+\frac{2^2}{5}+\frac{3^2}{7}+\dots +\frac{{1001}^2}{2003})\\ & =\frac{(2^2-1^2)}{3}+\frac{(3^2-2^2)}{5}+\frac{4^2-3^2}{7}+\dots +\frac{{1001}^2-{1000}^2}{2001}+1-\frac{{1001}^2}{2003}.\end{array}$$

Notice how etc. This is because all of these are in the form . There are of these terms since it begins at and ends at so Therefore, We can either manually calculate or notice that , so Therefore,

$$\begin{array}{rl}a-b& =\frac{(2^2-1^2)}{3}+\frac{(3^2-2^2)}{5}+\frac{4^2-3^2}{7}+\dots +\frac{{1001}^2-{1000}^2}{2001}+1-\frac{{1001}^2}{2003}\\ & =1001-\frac{{1001}^2}{2003}\\ & >1001-\frac{{1001}^2}{2002}\\ & =1001-\frac{1001}{2}\\ & =\frac{1001}{2}\\ & =500.5\end{array}$$

Since we can conclude that is closer to than

Thus, our answer is

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.