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1997 PMWC Problems/Problem T2

Revision as of 08:33, 9 October 2007 by 1=2 (talk | contribs) (New page: ==Problem== Evaluate <math>1(\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})</math> <math>+3(\dfrac{1...)
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Problem

Evaluate

$1(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$

$+3(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$

$+5(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$

$+7(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$

$+9(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})+11(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$

$+13(\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})+15(\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$

$+17(\dfrac{1}{9}+\dfrac{1}{10})+19(\dfrac{1}{10})$


Solution

We can group them:

$\dfrac{1}{1}+\dfrac{1+3}{2}+\dfrac{1+3+5}{3}+\cdots+\dfrac{1+3+5+7+9+11+13+15+17+19}{10}$

The sum of the first n odd numbers is n^2, so we can replace a lot of that stuff:

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


That's just $1+2+3+4+5+6+7+8+9+10=55$.

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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