If n be a positive integer, then prove by binomial theorem that the integral part of $(7+4\sqrt3)^n$ is an odd number. I wanted to know, how can i prove this?
If $n$ be a positive integer, then prove by binomial theorem that the integral part of $(7+4\sqrt3)^n$ is an odd number.

Consider the number $(7+4\sqrt{3})^n+(7-4\sqrt{3})^n$.  Imagine expanding each term using the Binomial Theorem, and adding. The terms that involve odd powers of $\sqrt{3}$ cancel, so we get an even integer $2K$.
Note that $(7-4\sqrt{3}$ is a small positive number, roughly $0.072$. So for any postive $n$, the number $(7-4\sqrt{3})^n$ is small positive, and in particular between $0$ and $0.072$. 
It follows that $(7+4\sqrt{3})^n$ is almost $2K$, but a little smaller. So the integer part of  $(7+4\sqrt{3})^n$ is $2K-1$.
Remark: When we have a question about $a+b\sqrt{d}$, where $a$, $b$, and $d$ are integers, with $d$ not a perfect square, the conjugate $a-b\sqrt{d}$ is likely to be useful. 
