Difference between revisions of "2006 Alabama ARML TST Problems/Problem 2"
(New page: ==Problem== Compute <math>\sum_{k=1}^{5}\sum_{n=1}^{6}kn</math>. ==Solution== For each value of k, the sum is equal to <math>1k+2k+3k+4k+5k+6k=21k</math>. <math>\sum_{k=1}^{5}21k=21*15=\...) |
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==See also== | ==See also== | ||
| + | {{ARML box|year=2006|state=Alabama|num-b=1|num-a=3}} | ||
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