If $x=1+\frac{3}{1!} \times \frac{1}{6}+\frac{3 \times 7}{2!}\left(\frac{1}{6}\right)^2+\frac{3 \times 7 \times 11}{3!}\left(\frac{1}{6}\right)^3+\ldots$,then $x^4$ equals

If $T_4$ represents the $4^{th}$ term in the expansion of $\left(5x + \frac{7}{x}\right)^{-3/2}$ and $x \notin \left[-\sqrt{\frac{7}{5}}, \sqrt{\frac{7}{5}}\right]$,then $\left(x^7 \sqrt{5x}\right) T_4 =$

To expand $(1 + 2x)^{-1/2}$ as an infinite series,the range of $x$ should be

If the coefficient of $x^{13}$ in the expansion of $\frac{(1+x)^2}{(1-2x)^3}$ is $A \times 2^{10}$,then $A=$

The binomial expansion $(7+3x)^{-2/5}$ is valid for all $x$ in the interval $\left(\frac{-7}{3}, \frac{7}{3}\right)$. If the $4^{th}$ term of its expansion is $kx^3$,then the value of $(7^{12/5}k)$ is:

Find a positive value of $m$ for which the coefficient of $x^{2}$ in the expansion $(1+x)^{m}$ is $6$.