# Binomial expansion- divisibility

1. Nov 7, 2014

### erisedk

1. The problem statement, all variables and given/known data
The integer next to (√3 + 1 )^2n is -- (n is a natural number)
Ans: Divisible by 2^(n+1)

2. Relevant equations

3. The attempt at a solution
(√3 + 1 )^2n will have an integral and a fractional part.
So, I + f = (√3 + 1 )^2n
(√3 - 1 )^2n will always be fractional as (√3 - 1) < 1
So, f' = (√3 - 1 )^2n

On adding I + f and f', all the irrational terms (which are also the even terms) will cancel out and the odd terms will get added.
I + f + f' = 2 [ (2nC0 √3^2n) + (2nC2 √3^2n-2) + .... ]
I + f + f' = 2 [ (2nC0 3^n) + (2nC2 3^n-1) + .... ]

Since I + f + f' is an even integer, 0<f<1, 0<f'<1 and 0<f+f'<2
f+f'=1

So, I = 2 [ (2nC0 3^n) + (2nC2 3^n-1) + .... ] - 1
I need the integer next to (√3 + 1 )^2n. Let the required integer be denoted by I'.

I' = 2[ (2nC0 3^n) + (2nC2 3^n-1) + .... ] - 1 + 1
I' =2 [ (2nC0 3^n) + (2nC2 3^n-1) + .... ]

From here, the only thing I infer is that I' is divisible by 2. How do I show that this is divisible by 2^(n+1) ?

2. Nov 7, 2014

### Ray Vickson

Several things are wrong with your post. First, do you mean $(\sqrt{3}+1)^{2n}$ or $(\sqrt{3}+1)^2 n$? I suspect you mean the first one, in which case you need parentheses, like this: (√3 + 1)^(2n). Second, do you mean that the answer you write is supposed to be true for all integers n ≥ 1? If that is what you mean the result is false. Or, do you mean you want to "solve for n" that produces the result cited? That is a very different type of question.