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