even number - binomial

Question

$$(1+\sqrt 2)^n+(1-\sqrt2)^n$$ if $$ n \in N $$ It's always an even number.

I tried to solve for the binomial, but I could not, any idea to be able to proceed $$(a+b)^n=\sum_{k=0}^n{n \choose k} a^{n-k}b^k$$

Answer 1

Applying the binomial expansion you gave,

$$(1+\sqrt2)^n=\sum_{k=0}^n{n \choose k} \sqrt2^k$$

$$(1-\sqrt2)^n=\sum_{k=0}^n{n \choose k} (-\sqrt2)^k$$

In their sum, terms of odd $k$ cancel, and terms of even $k$ are duplicated, so the sum is $$2\sum_{j=0}^{\lfloor\frac n2\rfloor}{n \choose 2j} \sqrt2^{2j},$$

which is clearly an even integer, because it's $2$ times an integer --

binomial coefficients and even non-negative powers of $\sqrt2$ are integers -- no induction needed.

Answer 2

For all $n \in \mathbb{N}$, let $u_n = \left( 1 +\sqrt{2} \right)^n + \left(1- \sqrt{2} \right)^n$.

First, for all $n$, one has $u_{n+2}=2u_{n+1} + u_n$ : indeed $$u_{n+2} = \left( 1 +\sqrt{2} \right)^{n+2} + \left(1- \sqrt{2} \right)^{n+2} = \left( 1 +\sqrt{2} \right)^2\left( 1 +\sqrt{2} \right)^n + \left( 1 -\sqrt{2} \right)^2\left(1- \sqrt{2} \right)^n $$

$$= (1+2(\sqrt{2}+1))\left( 1 +\sqrt{2} \right)^n + \left( 1 -2(\sqrt{2}+1) \right)\left(1- \sqrt{2} \right)^n$$

$$= \left( 1 +\sqrt{2} \right)^n + 2 \left( 1 +\sqrt{2} \right)^{n+1} + \left(1- \sqrt{2} \right)^n - 2\left(1- \sqrt{2} \right)^{n+1}$$

So you get $$u_{n+2}=2u_{n+1} + u_n$$

Now, a small induction shows you that $(u_n)$ is even for each $n$.

Answer 3

You can restore the difference equation with roots $1-\sqrt{2}$ and $1+\sqrt{2}$: $$x^2-2x-1=0.$$ Hence: $$a_n=2a_{n-1}+a_{n-2}, a_1=2,a_2=6,$$ which has a solution: $$a_n=(1-\sqrt{2})^n+(1+\sqrt{n})^n.$$ Obviously, it has all terms even.