Difference between revisions of "1997 PMWC Problems/Problem T2"
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That's just <math>1+2+3+4+5+6+7+8+9+10=\frac{10(10+1)}{2}=55</math>. | That's just <math>1+2+3+4+5+6+7+8+9+10=\frac{10(10+1)}{2}=55</math>. | ||
| − | ==See | + | ==See Also== |
{{PMWC box|year=1997|num-b=T1|num-a=T3}} | {{PMWC box|year=1997|num-b=T1|num-a=T3}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 15:06, 15 May 2012
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) (Error compiling LaTeX. Unknown error_msg)
Solution
We can group them:
![\[\dfrac{1}{1}+\dfrac{1+3}{2}+\dfrac{1+3+5}{3}+\cdots+\dfrac{1+3+5+7+9+\ldots+19}{10}\]](http://latex.artofproblemsolving.com/c/0/b/c0b555410b82987928acb86250badaa6f70e9827.png)
The sum of the first 
odd numbers is 
, so we can simplify:
![\[\dfrac{1}{1}+\dfrac{4}{2}+\dfrac{9}{3}+\cdots+\dfrac{100}{10}\]](http://latex.artofproblemsolving.com/c/3/6/c36b05197bfa3d13bcb2bba23c35757471ecc910.png)
That's just 
.
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 | ||
