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 = $

  • A
    $0$
  • B
    $1$
  • C
    $2$
  • D
    $0$ or $2$

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$(D)$ $g(2m, 2n) = (g(m, n))^2$ for all positive integers $m, n$

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The sum of the series $\sum\limits_{r = 0}^n {(-1)^r \, ^nC_r \left( \frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{2^{3r}} + \frac{15^r}{2^{4r}} + \dots + m \text{ terms} \right)}$ is

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The coefficient of $t^{24}$ in the expansion of $(1 + t^2)^{12}(1 + t^{12})(1 + t^{24})$ is

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