Question

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

• A. $evennumber$
• B. $oddnumber$
• C. $evenoranoddnumberdependinguponthevalueofn$
• D. $Noneofthis$

Answer 1

The correct option is B $oddnumber$

Consider the problem

Let the Binomial distribution theorem,

${\left(x+y\right)}^n=a_0{x}^n+a_1{x}^{n-1}y+a_2{x}^{n-2}{y}^2+.....a_{n-1}x{y}^{n-1}+a_n{y}^n$

And,

Also $n\ge 1,a_0=1$

$a_0{x}^n={1.7}^n=7^n$

First term is odd

Because, $7^n$ is odd for $n\ge 1$

And,

Exponent of $y$ can either be odd and even.

Now, we take.

Case I

Exponent is even.

$\begin{array}{c}{y}^{2m},m\in positiveintegers\\ \\ y^{2m}={\left(4\sqrt{3}\right)}^{2m}={\left(16.3\right)}^m={48}^m\end{array}$

Case II

Exponent is odd.

$\begin{array}{c}{y}^{2m+1},m\in positiveintegers\\ \\ y^{2m+1}={\left(4\sqrt{3}\right)}^{2m+1}={\left(4\sqrt{3}\right)}^{2m}.\left(4\sqrt{3}\right)={48}^m\left(4\sqrt{3}\right)\end{array}$

Conclusion :

${\left(x+y\right)}^n=\underset{odd}{a_0{x}^n}+\underset{non-\mathrm{int}eger}{a_1{x}^{n-1}y}+\underset{even}{a_2{x}^{n-2}{y}^2}+\underset{non-\mathrm{int}eger}{a_3{x}^{n-3}{y}^3}+\underset{even}{a_4{x}^{n-4}{y}^4}+........$

The integer part of our result will always be sum of one odd integers added to a even integers, which must result in a odd integer .

Hence, option $B$ is correct.