# (k+1)!

• Oct 10th 2007, 01:40 PM
(k+1)!
How is this evaluated. Is it k(k+1) or is it (k+1)(k+2). This is the problem if anyone wants to try it out.

Prove by induction on n that n!>2^n for all integers n such that n >or= 4
• Oct 10th 2007, 11:37 PM
DivideBy0
$(k+1)!=(k+1) \cdot k \cdot (k-1) \cdot ... \cdot 2 \cdot 1$

Proposition: $n!>2^n$ for $n\geq 4$.

Proof: We will prove the proposition by induction on a variable $n$.

If $n=4$, we have $4!>2^4$ or $24>16$, which is true.

Assume: $n! >2^n$ for $4\leq n \leq k$

Taking $n=k$, we have

$k!>2^k$

Multiplying by k+1, we have

$k!(k+1)>2^k(k+1)$ or $(k+1)!>2^k(k+1)$

And since $k \geq 4$, the minimum value we can have for $k+1$ is 5, so we have

$2^k(k+1)>2^k \cdot 2$ or $2^k(k+1)>2^{k+1}$

By transitivity, we have

$(k+1)!>2^{k+1}$

Hence, $(k+1)!>2^{k+1}$ for $n \geq 4$, QED