Solve the problem.
-Suppose you wish to prove the statement that follows using the extended principle of mathematical is $( n + 1 ) ! 3 ^ { n }$, for positive integers $n \geq 4$.

Assume that $S _ { k }$ is true for $k \geq 4$ where $S _ { k }$ is the statement $( n + 1 ) ! 3 ^ { n }$, and write the statement $S _ { k + 1 }$
A) $S _ { k + 1 } : ( k + 2 ) ! 3 ^ { k }$
B) $S _ { k + 1 } : ( k + 1 ) ! 4 ^ { k }$
C) $S _ { k + 1 } : ( k + 1 ) ! 3 ^ { k + 1 }$
D) $S _ { k + 1 } : ( k + 2 ) ! 3 ^ { k + 1 }$ ion.

Question

Solve the problem.
-Suppose you wish to prove the statement that follows using the extended principle of mathematical is $( n + 1 ) ! > 3 ^ { n }$, for positive integers $n \geq 4$.

Assume that $S _ { k }$ is true for $k \geq 4$ where $S _ { k }$ is the statement $( n + 1 ) ! > 3 ^ { n }$, and write the statement $S _ { k + 1 }$
A) $S _ { k + 1 } : ( k + 2 ) ! > 3 ^ { k }$
B) $S _ { k + 1 } : ( k + 1 ) ! > 4 ^ { k }$
C) $S _ { k + 1 } : ( k + 1 ) ! > 3 ^ { k + 1 }$
D) $S _ { k + 1 } : ( k + 2 ) ! > 3 ^ { k + 1 }$ ion.

Quizplus · Accepted Answer

The statement $S_{k+1}$ should increment both the factorial and the exponent by 1 to maintain the structure of the original statement, thus making $(k+2)! > 3^{k+1}$ the correct formulation.

Solve the Problem (n+1)!>3n( n + 1 ) ! > 3 ^ { n }(n+1)!>3n

Solve the Problem $( n + 1 ) ! > 3 ^ { n }$