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

(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...)
 
(fmtting)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
 
 
Evaluate
 
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>
+
<cmath>\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) \\
−
<math>+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})</math>
+
&+& 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)\\
−
<math>+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})</math>
+
&+&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)\\
−
<math>+7(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})</math>
+
&+&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>
−
<math>+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})</math>
 
−
 
 
−
<math>+13(\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})+15(\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})</math>
 
−
 
 
−
<math>+17(\dfrac{1}{9}+\dfrac{1}{10})+19(\dfrac{1}{10})</math>
 
−
 
 
  
 
==Solution==
 
==Solution==
−
 
 
We can group them:
 
We can group them:
  
−
<math>\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}</math>
+
<cmath>\dfrac{1}{1}+\dfrac{1+3}{2}+\dfrac{1+3+5}{3}+\cdots+\dfrac{1+3+5+7+9+\ldots+19}{10}</cmath>
  
−
The sum of the first n odd numbers is n^2, so we can replace a lot of that stuff:
+
The sum of the first <math>n</math> odd numbers is <math>n^2</math>, so we can simplify:
  
−
<math>\dfrac{1}{1}+\dfrac{4}{2}+\dfrac{9}{3}+\cdots+\dfrac{100}{10}</math>
+
<cmath>\dfrac{1}{1}+\dfrac{4}{2}+\dfrac{9}{3}+\cdots+\dfrac{100}{10}</cmath>
  
 +
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=55</math>.
+
==See also==
 +
{{PMWC box|year=1997|num-b=T1|num-a=T3}}
  
−
==See Also==
+
[[Category:Introductory Algebra Problems]]
−
 
 
−
{{PMWC box|year=1997|num-b=T1|num-a=T3}}
 

Revision as of 15:45, 9 October 2007

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:

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

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1997_PMWC_Problems/Problem_T2&oldid=17730"