# Difference between revisions of "1997 PMWC Problems/Problem T2"

## Problem

Evaluate

$\begin{eqnarray*} && 1 \left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right) \\ &+& 3\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\\ &+&5\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\\ &+&7\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\\ &+&9\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)+11\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\\ &+&13\left(\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)+15\left(\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\\ &+&17\left(\dfrac{1}{9}+\dfrac{1}{10}\right)+19\left(\dfrac{1}{10}\right)\end{eqnarray*}$

## Solution

We can group them:

$$\dfrac{1}{1}+\dfrac{1+3}{2}+\dfrac{1+3+5}{3}+\cdots+\dfrac{1+3+5+7+9+\ldots+19}{10}$$

The sum of the first $n$ odd numbers is $n^2$, so we can simplify:

$$\dfrac{1}{1}+\dfrac{4}{2}+\dfrac{9}{3}+\cdots+\dfrac{100}{10}$$

That's just $1+2+3+4+5+6+7+8+9+10=\frac{10(10+1)}{2}=55$.