First 10 digits after decimal point in the number $(1+\sqrt{3})^{2015}$

Question

The question is how to find first 10 digits after decimal point in the number $(1+\sqrt{3})^{2015}$.

I keep running into this kind of problems in a context of symmetric polynomials.

Answer 1

Hint: (i) The number $(1+\sqrt{3})^{2015}+(1-\sqrt{3})^{2015}$ is an integer. We can see this by imagining expanding using the Binomial Theorem. Terms involving odd powers of $\sqrt{3}$ cancel.

(ii) The number $(1-\sqrt{3})^{2015}$ is extremely small in absolute value, and negative. So $(1+\sqrt{3})^{2015}$ is nearly an integer.

Answer 2

Another way to see that $u_n=(1+\sqrt{2})^n+(1-\sqrt{2})^n$ is an integer is to exhibit a simple three-term recursion for it. $$ u_{n+1} = 2u_n+u_{n-1} $$