Binomial Theorem Question - None

Question

Integral part of $(7 + 4\sqrt{3})^n$ if $n \in \mathbb{N}$ is

Accepted Answer

Solution:

Here, $\forall n \in \mathbb{N}, (7 + 4\sqrt{3})^n otin \mathbb{N}$

Denote $(7 + 4\sqrt{3})^n$ by $I + f$, where $I$ is an integer and $f \in \mathbb{R}$ such that $0 < f < 1$

Since $0 < 7 - 4\sqrt{3} < 1$, we can denote $(7 - 4\sqrt{3})^n$ by $G$ where $G \in \mathbb{R}$ such that $0 < G < 1$

Now, $I + f = (7 + 4\sqrt{3})^n = 7^n + \binom{n}{1}7^{n-1}(4\sqrt{3}) + \binom{n}{2}7^{n-2}(4\sqrt{3})^2 + \ldots$

$G = (7 - 4\sqrt{3})^n = 7^n - \binom{n}{1}7^{n-1}(4\sqrt{3}) + \binom{n}{2}7^{n-2}(4\sqrt{3})^2 - \ldots$

Adding the two equations cancels irrational terms, giving $I + f + G = 2[7^n + \binom{n}{2}7^{n-2}(4\sqrt{3})^2 + \ldots]$, which is an even integer.

Since $0 < f + G < 2$, we have $I = 2k - 1$ (an odd number) for some integer $k$.

∴ Answer is (b)

Question: