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== | ||
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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]] |