If sum of the coefficient of the first, second and third terms of the expansion of ${\left( {{x^2} + \frac{1}{x}} \right)^m}$ is $46$, then the coefficient of the term that doesnot contain $x$ is :-

If the coefficients of $x$ and $x^{2}$ in the expansion of $(1+x)^{p}(1-x)^{q}, p, q \leq 15$, are $-3$ and $-5$ respectively, then the coefficient of $x ^{3}$ is equal to $............$

Expand using Binomial Theorem $\left(1+\frac{ x }{2}-\frac{2}{ x }\right)^{4}, x \neq 0$

Find $a$ if the $17^{\text {th }}$ and $18^{\text {th }}$ terms of the expansion ${(2 + a)^{{\rm{50 }}}}$ are equal.

For a positive integer $n,\left(1+\frac{1}{x}\right)^{n}$ is expanded in increasing powers of $x$. If three consecutive coefficients in this expansion are in the ratio, $2: 5: 12,$ then $n$ is equal to

The term independent of $x$ in the expansion of $\left( {\frac{1}{{60}} - \frac{{{x^8}}}{{81}}} \right).{\left( {2{x^2} - \frac{3}{{{x^2}}}} \right)^6}$ is equal to