1997 PMWC Problems/Problem T2

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Problem

Evaluate

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

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

1997 PMWC (Problems)
Preceded by Problem T1	Followed by Problem T3
I: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 T: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10

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1997 PMWC Problems/Problem T2

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Solution

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