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

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&+&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)\\
 
&+&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)\\
 
&+&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)</cmath>
+
&+&17\left(\dfrac{1}{9}+\dfrac{1}{10}\right)+19\left(\dfrac{1}{10}\right)\end{eqnarray*}</cmath>
  
 
==Solution==
 
==Solution==

Latest revision as of 19:07, 10 March 2015

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

See Also

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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