1+3+5+7+9+...+m = n²

m=2n-1

function SumaImpares(n: Natural) is
  m = 2·n - 1
  1+3+5+7+9+...+m = n²
  
if n=1 then
  1 = 1² = 1
  is True

if s(n) then s(n+1)
  1+3+5+7+9+...+m + 2(n+1)-1 = (n+1)²
  n² + 2n+2 - 1 = (n+1)²
  n² + 2n+2 - 1 = n²+2n+1
  n² + 2n + 1 = n² + 2n + 1 
  is True

n = 1  ⟹  s = 1
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n = 2  ⟹  s = 4    ∆n = 3 = 2 + 1 = n + n-1 = 2n-1
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n = 3  ⟹  s = 9    ∆n = 5 = 2 + 1 = n + n-1 = 2n-1
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n = 4  ⟹  s = 16    ∆n = 7 = 4 + 3 = n + n-1 = 2n-1
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n = 5  ⟹  s = 25    ∆n = 9 = 5 + 4 = n + n-1 = 2n-1
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