Question

If $n$ is a positive even integer, which of the following will always be integers

• A. $(\sqrt{2}+1{)}^n+(\sqrt{2}-1{)}^n$
• B. $(\sqrt{2}+1{)}^n-(\sqrt{2}-1{)}^n$
• C. $(\sqrt{2}+1{)}^{2n+1}+(\sqrt{2}-1{)}^{2n+1}$
• D. $(\sqrt{2}+1{)}^{2n+1}-(\sqrt{2}-1{)}^{2n+1}$

Answer 1

Answer: (A), (D)

The correct options are
A $(\sqrt{2}+1{)}^n+(\sqrt{2}-1{)}^n$
D $(\sqrt{2}+1{)}^{2n+1}-(\sqrt{2}-1{)}^{2n+1}$

We know that
${(x+a)}^n+{(x-a)}^n=2{(}^n{C}_0{x}^n{+}^n{C}_2{x}^{n-2}{a}^2+....)$
${(x+a)}^n-{(x-a)}^n=2{(}^n{C}_1{x}^{n-1}a{+}^n{C}_3{x}^{n-3}{a}^3+....)$
Now consider option $A,$
$(\sqrt{2}+1{)}^n+(\sqrt{2}-1{)}^n$
$=2{[}^n{C}_0\left(\sqrt{2}{)}^n{+}^n{C}_2\right(\sqrt{2}{)}^{n-2}+....]$
Since, $n$ is a positive even integer. so, $n-2,n-4,....$ are all positive even integers.
So, $(\sqrt{2}{)}^{\text{even no.}}=\text{integer}$
Hence, $(\sqrt{2}+1{)}^n+(\sqrt{2}-1{)}^n$ is always an integer.
Option $B,$
$(\sqrt{2}+1{)}^n-(\sqrt{2}-1{)}^n$
$=2{[}^n{C}_1{x}^{n-1}{+}^n{C}_3{x}^{n-3}+....]$
Since, $n$ is a positive even integer. so, $n-1,n-3,....$ are all positive odd integers.
So, $(\sqrt{2}{)}^{\text{odd no.}}=\text{irrational}$
Hence,$(\sqrt{2}+1{)}^n-(\sqrt{2}-1{)}^n$ is not an integer.
Option $C,$
$(\sqrt{2}+1{)}^{2n+1}+(\sqrt{2}-1{)}^{2n+1}$
$=2{[}^{2n+1}{C}_0\left(\sqrt{2}{)}^{2n+1}{+}^{2n+1}{C}_2\right(\sqrt{2}{)}^{2n-1}+....]$
Since, $n$ is a positive even integer. So, $2n+1,2n-1,....$ are all positive odd integers.
So, $(\sqrt{2}{)}^{\text{odd no.}}=\text{irrational}$
Hence, $(\sqrt{2}+1{)}^{2n+1}+(\sqrt{2}-1{)}^{2n+1}$ is not an integer.
Option $D,$
$(\sqrt{2}+1{)}^{2n+1}-(\sqrt{2}-1{)}^{2n+1}$
$=2{[}^{2n+1}{C}_1{x}^{2n}{+}^{2n+1}{C}_3{x}^{2n-2}{a}^3+....]$
Since, $n$ is a positive even integer. so, 2n,2n-2 ,....are all positive even integers.
So, $(\sqrt{2}{)}^{\text{even no.}}=\text{integer}$
Hence,$(\sqrt{2}+1{)}^{2n+1}-(\sqrt{2}-1{)}^{2n+1}$ is always an integer.