Considere a prova do seguinte teorema, utilizando o Princípio da Indução Finita: Para todo n element of straight natural numbers, 2 plus 5 plus 8 plus... plus left parenthesis 2 plus 3 n right parenthesis equals fraction numerator left parenthesis n plus 1 right parenthesis left parenthesis 4 plus 3 n right parenthesis over denominator 2 end fraction.
Assinale a alternativa que corresponde à hipótese de indução. a.
Fixado k element of straight natural numbers, 2 plus 5 plus 8 plus... plus left parenthesis 2 plus 3 k right parenthesis equals fraction numerator 3 k over denominator 2 end fraction. b.
Fixado k element of straight natural numbers, 2 plus 5 plus 8 plus... plus left parenthesis 2 plus 3 left parenthesis k plus 1 right parenthesis right parenthesis equals fraction numerator left parenthesis k plus 2 right parenthesis left parenthesis 4 plus 3 left parenthesis k plus 1 right parenthesis right parenthesis over denominator 2 end fraction. c.
Fixado k element of straight natural numbers, 2 plus 5 plus 8 plus... plus left parenthesis 2 plus 3 k right parenthesis equals fraction numerator left parenthesis k plus 1 right parenthesis left parenthesis 4 plus 3 k right parenthesis over denominator 2 end fraction. d.
2 plus 5 equals fraction numerator left parenthesis 1 plus 1 right parenthesis left parenthesis 4 plus 3.1 right parenthesis over denominator 2 end fraction. e.
Fixado k element of straight natural numbers, 2 plus 5 plus 8 plus... plus left parenthesis 2 plus 3 k right parenthesis equals fraction numerator k left parenthesis 4 plus 3 k right parenthesis over denominator 2 end fraction.
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Resposta:
Fixado k E N, 2 + 5 + 8 +... +(2 plus 3 k )= (k+1)(4+3k)/2