# Prove this inequality with binomial

1. Nov 17, 2011

### Mathitalian

1. The problem statement, all variables and given/known data
Prove that

$\sum_{k=0}^n {3k\choose k}\ge \frac{5^n-1}{4}$

2. Relevant equations

${3k\choose k}= \frac{(3k)!}{k!(2k)!}$

3. The attempt at a solution

I tried using the induction principle, but...

Here my attempt:

For $n=0$ 1>0 ok

Suppose that is true for $n$, i.e.:

$\sum_{k=0}^n {3k\choose k}\ge \frac{5^n-1}{4}$

Now:

$\sum_{k=0}^{n+1} {3k\choose k}= \sum_{k=0}^{n} {3k\choose k}+ {3(n+1)\choose (n+1)}\ge \frac{5^n-1}{4}+{3(n+1)\choose (n+1)}$

But now I don't know what to do, maybe it is not the correct way to show this... I need your help

2. Nov 19, 2011

### Mathitalian

Please, can someone help me? I think that it exists a different way to prove this inequality, but I don't know how to proceed :(

3. Nov 19, 2011

### Dick

What you actually need to show is that C(3(n+1),(n+1))>=(5^(n+1)-1)/4-(5^n-1)/4. Do you see why? You can definitely simplify the right side a lot. Now look at both sides for small values of n. Can you see how to continue? Think about it. I haven't much beyond this. Help me!

Last edited: Nov 19, 2011