prove that ${2n \choose n} \le 4^{n}$

Question

i need to prove that ${2n \choose n} \le 4^{n}$ for all $ n \in N $.

I tried to prove it that way: $$ 4^{n} = (1+1)^{2n} = \sum_{i=0}^{2n} {2n \choose i}1^{2n-i}1^{i} = \sum_{i=0}^{2n} {2n \choose i} $$

We can see that ${2n \choose n}$ is part of the sum $\sum_{i=0}^{2n} {2n \choose i} $ when $i=n$ and the other members of the sum are greater than zero.

So we got that: $$ {2n \choose n} \lt 4^{n} $$

Is this enough to prove what I asked for? because I didn't show how we can get that $ {2n \choose n} = 4^{n} $, I found that the this happens when $n=0$ but 0 is not a neutral number.

Answer 1

When you are asked to show

$$a_n\leq b_n$$

you need to show that, for each $n$, $a_n<b_n$ or $a_n=b_n$. You have shown that $a_n<b_n$ is true for all $n>0$, so necessarily $a_n\leq b_n$. Thus you are done.

Answer 2

Your proof is valid and shorter than the common way of induction.

$0$ is not a neutral number.

For the equality, yes, LHS is equal to RHS when $n = 0$. But according to some sources, $0$ is a natural number although in some sources it is not.

Answer 3

Your proof of the inequality is valid; you've expressed $4^n$ as the sum of entries in a row of Pascal's triangle, with $\binom{2n}{n}$ in the middle. If you want equality, you need your row to only have its central element; but it has $n+1$ elements, so $n=0$ is the unique option.