An Important Theorem - Practice Questions & MCQ

Finding the nature of an integral part of the expression.
If the given expansion is in the form of $N=(a+\sqrt{b})^n \quad(n \in N)$
Working rule:
Step 1: Choose $\mathrm{N}^{\prime}=(\mathrm{a}-\sqrt{\mathrm{b}})^{\mathrm{n}}$ or $(\sqrt{\mathrm{b}}-\mathrm{a})^{\mathrm{n}}$ according as $\mathrm{a}>\sqrt{\mathrm{b}}$ or $\sqrt{\mathrm{b}}>\mathrm{a}$
Step 2: Use N + N' or $\mathrm{N}-\mathrm{N}$ ' such that result is an integer
I.e. $(a+\sqrt{b})^n+(a-\sqrt{b})^n$ or $(a+\sqrt{b})^n-(a-\sqrt{b})^n$ is an integer

Step 3: Now use the concept of greatest integer function and fractional part of a function, $N=I+f$, where I am an integral part of $N$ i.e., [ $N$ ] and $f$ is a fractional part of $N$, i.e. \{ $N$ \}.
For example, the integral part of $P=(3 \sqrt{3}+5)^{2 n+1}(n \in N)$ is an even number.
Now consider, $P^{\prime}=(3 \sqrt{3}-5)^{2 n+1}$ here $0<P^{\prime}<1$
Use, $P-P^{\prime}=2\left[{ }^{2 n+1} C_1(3 \sqrt{3})^{2 n} 5^1+{ }^{2 n+1} C_3(3 \sqrt{3})^{2 n-2}(5)^3+\ldots \ldots\right]$
$I+f-P^{\prime}=2 k(k \in N) \quad(\mathrm{P}=\mathrm{I}+\mathrm{f})$
$-1<f-P^{\prime}<1$ but $f-P^{\prime}$ is an integer $\Rightarrow f-P^{\prime}=0 \Rightarrow I=2 k$
Hence, integral part of $P=(3 \sqrt{3}+5)^{2 n+1}(n \in N)$ is an even integer