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  • 24-04-2018
  • Mathematics
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Аноним Аноним
  • 24-04-2018
[tex] $\sum\limits_{n=1}^{15} (2n-1)[/tex]

When n = 1, 2n - 1 = 2(1) - 1 = 1
When n = 2, 2n - 1 = 2(2) - 1 = 3
When n = 3, 2n - 1 = 2(3) - 1 = 5
...

So: the total will be :
[tex]= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 [/tex]
[tex]= 225 [/tex]

[tex]$\sum\limits_{n=1}^{15} (2n-1) = 225 [/tex]
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