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.
 A: 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. 
A: I'm working with the hypothesis that "integral part" means "floor".
$$\begin{align}
(7+4\sqrt{3})^n + (7 - 4\sqrt{3})^n &= \sum_{k=0}^n \binom{n}{k}7^{n-k}4^k\sqrt{3}^k + \sum_{k=0}^n \binom{n}{k}7^{n-k}(-1)^k4^k\sqrt{3}^k\\
&= 2\sum_{m=0}^{\lfloor n/2\rfloor} \binom{n}{2m}7^{n-2m}4^{2m}3^m
\end{align}$$
is even. $0 < 7 - 4\sqrt{3} < 1$. Thus the integral part of $(7+4\sqrt{3})^n$ is
$(7+4\sqrt{3})^n + (7-4\sqrt{3})^n - 1$, an odd number.
A: First do the obvious, and see what the binomial theorem says when applied to $\left(7+4\sqrt3\right)^n$:
$$\left(7+4\sqrt3\right)^n=\sum_{k=0}^n\binom{n}k7^{n-k}\left(4\sqrt3\right)^k\;.$$
A term $\dbinom{n}k7^{n-k}\cdot\left(4\sqrt3\right)^k$ is an integer if and only if $k$ is even, and $\left(4\sqrt3\right)^2=48$, so the integer part is
$$\sum_{k\ge 0}\binom{n}k\left(7^{n-2k}\cdot48^k\right)\;.$$
Each of the terms in this last sum is even except the $k=0$ term; why? And if exactly one term is odd, then the sum itself must be ... ?
Note: I’m interpreting the integer part to be the integer $a$ when the power is written in the form $a+b\sqrt3$ with $a,b\in\Bbb Z$. If instead $\left\lfloor\left(7+4\sqrt3\right)^n\right\rfloor$ is intended, see the answers given by Daniel and André.
A: I'm working under the assumption that with "integral part" you mean the part of the answer that is an integer.
1) Let's take the binomial theorem, which gives us:
$(x + y)^n = a_0x^n + a_1x^{n-1}y + a_2x^{n-2}y^2 + ... + a_{n-1}xy^{n-1} + a_{n}y^n$
From the initial conditions $\rightarrow n\ge1, a_0 = 1$
$\rightarrow a_{0}x^n = 1 \dot\ 7^n = 7^n$
$\rightarrow$ first term is odd, because $7^n$ is odd for $n\ge1$.
2) The exponent for the $y$ can either be odd or even.
2a) Case 1 - Exponent is even: $y^{2m}, m\in positive\ Integers$
$y^{2m} = (4\sqrt{3})^{2m} = (16\dot\ 3)^m = 48^m$
$\rightarrow$ Any term with an even exponent for $y$ (the term with the squareroot) will result in an integer term. It will always be an even integer, because it results from a product with an even number.
2b) Case 2 - Exponent is odd: $y^{2m+1}, m\in positive\ Integers$
$y^{2m+1} = (4\sqrt{3})^{2m+1} = (4\sqrt{3})^{2m}\dot\ (4\sqrt{3}) = 48^m(4\sqrt{3}) $
$\rightarrow$ An odd exponent for $y$ (the term with the squareroot) will always lead to a non-integer term.
Conclusion: 
$(x + y)^n = \underbrace{a_0x^n}_{odd} + \underbrace{a_1x^{n-1}y}_{non-integer} + \underbrace{a_2x^{n-2}y^2}_{even} + \underbrace{a_3x^{n-3}y^3}_{non-integer} + \underbrace{a_4x^{n-4}y^4}_{even} + ... $
The integer part of our result will always be a sum of one odd integer added to a sum of even integers, which must always result in an odd integer.
