Question

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.

Answer 1

Resposta:

Fixado k E N, 2 + 5 + 8 +... +(2 plus 3 k )= (k+1)(4+3k)/2