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Given the properties of the function ( F(x) ), we can start by calculating ( F(1) = 1 ). Using the recursive relation ( F(x+1) = F(x) + 3x(x+1) + 1 ), we can compute the subsequent values:

  1. ( F(2) = F(1) + 3 \cdot 1 \cdot 2 + 1 = 1 + 6 + 1 = 8 )
  2. ( F(3) = F(2) + 3 \cdot 2 \cdot 3 + 1 = 8 + 18 + 1 = 27 )
  3. ( F(4) = F(3) + 3 \cdot 3 \cdot 4 + 1 = 27 + 36 + 1 = 64 )
  4. ( F(5) = F(4) + 3 \cdot 4 \cdot 5 + 1 = 64 + 60 + 1 = 125 )
  5. ( F(6) = F(5) + 3 \cdot 5 \cdot 6 + 1 = 125 + 90 + 1 = 216 )

The values are ( F(2) = 8 ), ( F(3) = 27 ), ( F(4) = 64 ), ( F(5) = 125 ), and ( F(6) = 216 ). The function exhibits a polynomial growth pattern, possibly resembling ( F(n) = n^3 ) as the outputs correspond to cubes of natural numbers.

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2w ago

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