induction method for sqrt

Question

$$(1+\sqrt 2)^n-(1-\sqrt2)^n=b_n\sqrt2$$ $$n\in N $$ $$b_n\in N$$ I needed to solve this by the induction method. The base and thesis of induction is easy, is to substitute n for n + 1, but I can not reduce the expression, someone could give an aid

Answer 1

Just write $(1+\sqrt{2})^{n+1} = (1+\sqrt{2})(1+\sqrt{2})^n$

You can then develop, while keeping the part which is to the $n$, for the induction. Do that on the other term also, and it should work

Answer 2

Since $\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}$, and $\left(1-\sqrt{2}\right)^2=3-2\sqrt{2}$, it is not too difficult to see that both $1+\sqrt{2}$ and $1-\sqrt{2}$ satisfy $$ x^2=2x+1\tag1 $$ This can be used to get a recurrence for $b_n$.

Answer 3

Might be easier to prove two statements.

That $(1 + \sqrt 2)^n - (1-\sqrt 2)^n = b_n \sqrt 2$

And $(1 + \sqrt 2)^n + (1 - \sqrt 2)^n = c_n$. where $b_n, c_n$ are integers.

Base cases:

For $n =1$ then $(1 + \sqrt 2)-(1-\sqrt 2) = 2\sqrt 2$ and $(1+\sqrt 2)+(1-\sqrt 2)=2$.

Induction cases:

If $(1 + \sqrt 2)^n -(1-\sqrt 2)^n = b_n \sqrt 2$ and $(1 + \sqrt 2)^n +(1-\sqrt 2)^n=c_n$

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

$(1 + \sqrt 2)^n(1 + \sqrt 2) - (1 - \sqrt 2)^n(1 - \sqrt 2) =$

$(1 + \sqrt 2)^n + \sqrt 2(1+ \sqrt 2)^n - (1-\sqrt 2)^n + (1-\sqrt 2)^n\sqrt 2 =$

$(1 + \sqrt 2)^n - (1-\sqrt 2)^n + \sqrt 2(1+ \sqrt 2)^n+ (1-\sqrt 2)^n\sqrt 2 =$

$b_n \sqrt 2 + c_n\sqrt 2 = (b_n + c_n) \sqrt 2$

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

$(1 + \sqrt 2)^n(1 + \sqrt 2) + (1 - \sqrt 2)^n(1 - \sqrt 2) =$

$(1 + \sqrt 2)^n + \sqrt 2(1+ \sqrt 2)^n + (1-\sqrt 2)^n - (1-\sqrt 2)^n\sqrt 2 =$

$(1 + \sqrt 2)^n + (1-\sqrt 2)^n + \sqrt 2(1+ \sqrt 2)^n - (1-\sqrt 2)^n\sqrt 2 =$

$c_n - b_n\sqrt 2\sqrt 2 = c_n - 2b_n$.

Although the best way to do this is with that binomial theorem.

$(1 \pm \sqrt 2)^n = \sum_{k=0}^n (\pm 1)^n {n\choose k}\sqrt 2^k$.

If $k$ is even then $(-1)^k =1$ so for

$(1+\sqrt2)^n - (1-\sqrt 2)^n =\sum_{k=0}^n (1-(-1)^k) {n\choose k}\sqrt 2^k$

the even terms will vanish.

If $k$ is odd then $(1-(-1)^k) = 2$ and $\sqrt 2^k = 2^{\frac {k-1}2}\sqrt 2$.

So

$(1+\sqrt2)^n - (1-\sqrt 2)^n =\sum_{k=0}^n (1-(-1)^k) {n\choose k}\sqrt 2^k=$

$\sum_{k=0; k \text{ odd}}^n 2{n\choose k}2^{\frac {k-1}2}\sqrt 2=$

$\sqrt 2(\sum_{k=0; k \text{ odd}}^n 2{n\choose k}2^{\frac {k-1}2}) =$

$\sqrt 2 b_n$ if we set $b_n$ equal to $\sum_{k=0; k \text{ odd}}^n 2{n\choose k}2^{\frac {k-1}2}$ which is an integer.