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What is the sum of the first 40 positive odd integers?

Guest Dec 5, 2018
 #1
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\(\sum \limits_{k=1}^{40}~2k-1 = \\ \left(2\sum \limits_{k=1}^{40}~k\right) - 40 = \\ 40\cdot 41-40 = 40^2 = 1600\)

Rom  Dec 5, 2018
edited by Rom  Dec 5, 2018
 #2
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The sum of the first N odd numbers is simply N^2 =40^2 =1,600

Guest Dec 6, 2018

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