How did my colleague conclude on equation (ii)?

[chwala]

Homework Statement

prove that,

$(n+1)!≥ 2^n. n$ when $n≥3$

Relevant Equations

mathematics induction

this is a solution posted by my colleague, i have a problem in understanding how he got to conclude on equation (ii)

is this not supposed to be $(k+2)!≥ 2^{k+1} (k+1)$
$(k+2)(k+1)!≥ 2^{k+1} (k+1)$ ?...

 

[etotheipi]

Assume that $(k+1)! \geq 2^k k$. Then notice that $(k+2)! = (k+2)(k+1)!$, and since $(k+1)! \geq 2^k k$ by assumption, you have $(k+2)! = (k+2)(k+1)! \geq (k+2)2^k k$.

You're trying to show that $(k+2)!$ is greater than $2^{k+1}(k+1)$ given that it holds for the case below, so you can't assume what's meant to be proven!

 

[PeroK]

Homework Statement:: prove that,

$(n+1)!≥ 2^n. n$ when $n≥3$
Relevant Equations:: mathematics induction

View attachment 269193
this is a solution posted by my colleague, i have a problem in understanding how he got to conclude on equation (ii)
View attachment 269194
is this not supposed to be $(k+2)!≥ 2^{k+1} (k+1)$
$(k+2)(k+1)!≥ 2^{k+1} (k+1)$ ?...

The inductive assumption is that $(k+1)!≥ 2^{k} (k)$, hence:
$$(k+2)! = (k+2)(k+1)! \ge (k+2)2^k (k)$$

 

[chwala]

aaaaaaah i have seen it...

 

[PeroK]

There must be a neater way to do this. That proof you attached is quite laboured.

 

[chwala]

There must be a neater way to do this. That proof you attached is quite laboured.

you mind showing a neater way now that i am conversant with how the solution was found...

 

[etotheipi]

You just need to state that $k(k+2) > 2(k+1)$ for all $k \geq 3$. Then it follows that$$(k+2)! \geq k(k+2)2^k > 2(k+1) 2^k = (k+1)2^{k+1}$$

 

[PeroK]

For the inductive step, all that is required is:
$$(k+2)! = (k+2)(k+1)! \ge (k+2)2^k(k) > (k+1)2^{k+1} \ \ \ (\text{as} \ k + 2 > k+1 \ \text{and} \ k > 2)$$