The fractional part of a real number $x$ is defined as $x - [x]$,where $[x]$ 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 $1.8$

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If the $6^{th}$ term in the expansion of the binomial $[\sqrt{2^{\log(10 - 3^x)}} + \sqrt[5]{2^{(x - 2)\log 3}}]^m$ is equal to $21$ and it is known that the binomial coefficients of the $2^{nd}$,$3^{rd}$ and $4^{th}$ terms in the expansion represent respectively the first,third and fifth terms of an $A.P.$ (the symbol $\log$ stands for logarithm to the base $10$),then $x = $

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Let $a > b > 0$ and $f(n) = a^{1/n} - b^{1/n}$, $J(n) = (a - b)^{1/n}$ for all $n \geq 2$. Then:

For non-negative integers $s$ and $r$,let $\binom{s}{r} = \begin{cases} \frac{s!}{r!(s-r)!} & \text{if } r \leq s \\ 0 & \text{if } r > s \end{cases}$. For positive integers $m$ and $n$,let $g(m, n) = \sum_{p=0}^{m+n} \frac{f(m, n, p)}{\binom{n+p}{p}}$,where for any non-negative integer $p$,$f(m, n, p) = \sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{p} \binom{p+n}{p-i}$. Then which of the following statements is/are $TRUE$?
$(A)$ $g(m, n) = g(n, m)$ for all positive integers $m, n$
$(B)$ $g(m, n+1) = g(m+1, n)$ for all positive integers $m, n$
$(C)$ $g(2m, 2n) = 2g(m, n)$ for all positive integers $m, n$
$(D)$ $g(2m, 2n) = (g(m, n))^2$ for all positive integers $m, n$

Ten trucks,numbered $1$ to $10$,are carrying packets of sugar. Each packet weighs either $999 \ g$ or $1000 \ g$,and each truck carries only packets of equal weight. The combined weight of $1$ packet selected from the first truck,$2$ packets from the second,$4$ packets from the third,and so on,up to $2^9$ packets from the tenth truck is $1022870 \ g$. Which trucks are carrying the lighter bags?

Let $(1 + x)(1 + x + x^2)(1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^{30}) = a_0 + a_1x + a_2x^2 + \dots + a_{465}x^{465}$. Then the sum $a_0 + a_2 + a_4 + \dots$ is equal to:

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