Revision history of "2006 Alabama ARML TST Problems/Problem 2"

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  • (cur | prev) 13:21, 2 March 20081=2 (talk | contribs). . (272 bytes) (+53)
  • (cur | prev) 13:21, 2 March 20081=2 (talk | contribs). . (219 bytes) (+219). . (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=\...)