Question

The fractional part of a real number x is $x-\left[x\right]$, where $\left[x\right]$ is the greatest integer less than or equal to x. Let $F_1$ and $F_2$ be the fractional parts of $(44-\sqrt{2017}{)}^{2017}$ and $(44+\sqrt{2017}{)}^{2017}$ respectively. Then $F_1+F_2$ lies between the numbers.

• A. $0$ and $0.45$
• B. $0.45$ and $0.9$
• C. $0.9$ and $1.35$
• D. $1.35$ and $1.8$

Answer 1

The correct option is C $0.9$ and $1.35$

$(44-\sqrt{2017}{)}^{2017}={44}^{2017}{}^{2017}{C}_0-{44}^{2016}{}^{2017}{C}_1|\sqrt{2017}|+{44}^{2015}{}^{2017}{C}_2(\sqrt{2017}{)}^2-{44}^{2014}{}^{2017}{C}_3(\sqrt{2017}{)}^3+...$

So, only odd powers of $\sqrt{2017}$ have fractional part and other terms have zero fractional part.

Let sum of odd powers is m.abc where $m=\left[x\right]$ and abc is $\left\{x\right\}$. Also let sum of even powers in $n$ which is positive integer. Then $(44-\sqrt{2017}{)}^{2017}=n-m.abc$

$=n-m-0.abc$

$=\underset{integerpart\left[x\right]}{\underset{}{(n-m-1)}}+\underset{fractionalpart\left\{x\right\}}{\underset{}{(1-0.abc)}}$

Also, $(44+\sqrt{2017}{)}^{2017}=n+m.abc=\underset{integerpart}{\underset{}{(n+m)}}+\underset{fractionalpart}{\underset{}{\left(0.abc\right)}}$

So, $F_1+F_2=(1-0.abc)+\left(0.abc\right)=1$

Thus, the answer is option (C).