Question

If $\left[x\right]$ denotes the greatest integer less than or equal to $x,$ then $\left[{(6\sqrt{6}+14)}^{2n+1}\right]$

• A. is an even integer
• B. is an odd integer
• C. depends on $n$
• D. none of these

Answer 1

The correct option is A is an even integer

$I+f={(6\sqrt{6}+14)}^{2n+1}$

Assuming $f^{{}^{\prime }}={(6\sqrt{6}-14)}^{2n+1}$..........(1)

Now ${I+f-f}^{{}^{\prime }}={(6\sqrt{6}+14)}^{2n+1}+{(6\sqrt{6}-14)}^{2n+1}$

$\Rightarrow {I+f-f}^{{}^{\prime }}=2[{{}^{2n+1}C}_1{\left(6\sqrt{6}\right)}^{2n}14+{{}^{2n+1}C}_3{\left(6\sqrt{6}\right)}^{2n+2}{14}^3+...]$

${I+f-f}^{{}^{\prime }}=2\left\{integer\right\}$ = even..............(2)

Now $0\le f\le 1$

Also $0\le f-f^{{}^{\prime }}\le 1$

Therefore using (1) we get, $0\le f-f^{{}^{\prime }}<0$ $f-f^{{}^{\prime }}=0$

Substituting respective values in (2) we get $I=$integer