By M.I prove that?

Question

$(2n)! < 2^{2n} (n!)^2$; How to prove this by mathematical induction ?

Are you in general familiar with the concept of mathematical induction? — mrtaurho, Oct 14 '18 at 19:17
In the First step we have to prove the basis , for, n = 1. Step 2 , we assume n = k' and then instep 3we prove our assumption withn = k + 1`. — user83869, Oct 14 '18 at 19:21

score 2 · Accepted Answer · answered Oct 14 '18 at 19:32

Start with the base case $n=1$ to get

$$\begin{align} (2\cdot 1)!<2^{2\cdot1}(1)!^2\Leftrightarrow 2<4 \end{align}$$

now assume the two cases $n=m$ and $n=m+1$ as following

$$\begin{align} (2\cdot m)!&<2^{2\cdot m}(m)!^2\\ (2\cdot (m+1))!&<2^{2\cdot (m+1)}(m+1)!^2 \end{align}$$

now observe that $(2m+2)(2m+1)(2m)!=(2m+2)!$ and that $(2m+2)^22^{2m}m!=2^{2\cdot (m+1)}(m+1)!^2$ to get from the first to the second case. Now we have to compare the factors of the RHS and the LHS and show which one is greater for $m\ge1$. So we get

$$\begin{align} (2(m+1))!=(2m+2)(2m+1)(2\cdot m)!&<(2m+2)(2m+1)2^{2\cdot m}(m)!^2\\ &<(2m+2)(2m+2)2^{2\cdot m}(m)!^2\\ &=2^2(m+1)^22^{2m}(m)!^2\\ &=2^{2(m+1)}(m+1)!^2\\ \Leftrightarrow (2(m+1))!&<2^{2(m+1)}(m+1)!^2 \end{align}$$

and we are done. For inductive proof concerning inequalities it is always a good to try to just multiply one side with the missing factor or to add the missing summand to one side. Then start to reshape it in way that the factor/summand of the other side shows up while always staying focused on the relation.

score 0 · Answer 2 · answered Oct 14 '18 at 19:20

0

Hint: $(2(n+1))!=(2n+2)!=(2n)!(2n+1)(2n+2)<$

$<(2n)!(2n+2)(2n+2)=(2n)!2^2(n+1)^2$

From here try to use the induction hypothesis.

answered Oct 14 '18 at 19:20

Mark

43,582

Isn't $(2n + 2)! = (2n)(2n + 1)(2n + 2)$ ? – user83869 Oct 14 '18 at 19:33
No, it is $(2n)!(2n+1)(2n+2)$. Try to check by the definition of factorial. – Mark Oct 14 '18 at 19:35

By M.I prove that?

2 Answers2