Write the general term in the expansion of $\left(x^{2}-y x\right)^{12}, x \neq 0$
It is known that the general term ${T_{r + 1}}{\rm{ \{ }}$ which is the ${(r + 1)^{{\rm{th }}}}$ term $\} $ in the binomial expansion of $(a+b)^{n}$ is given by ${T_{r + 1}} = {\,^n}{C_r}{a^{n - r}}{b^r}$
Thus, the general term in the expansion of $\left(x^{2}-y x\right)^{12}$ is
${T_{r + 1}} = {\,^{12}}{C_r}{\left( {{x^2}} \right)^{12 - r}}{( - yx)^r} = {( - 1)^r}{\,^{12}}{C_r} \cdot {x^{24 - 2r}}{y^r} = {( - 1)^r}{\,^{12}}{C_r} \cdot {x^{24 - r}} \cdot {y^r}$
If the coefficients of $x^2$ and $x^3$ are both zero, in the expansion of the expression $(1 + ax + bx^2) (1 -3x)^{t5}$ in powers of $x$, then the ordered pair $(a, b)$ is equal to
The expression $[x + (x^3-1)^{1/2}]^5 + [x - (x^3-1)^{1/2}]^5$ is a polynomial of degree :
The term independent of ' $x$ ' in the expansion of $\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}$, where $x \neq 0,1$ is equal to $.....$
Find the middle terms in the expansions of $\left(3-\frac{x^{3}}{6}\right)^{7}$
If the coefficients of the three successive terms in the binomial expansion of $(1 + x)^n$ are in the ratio $1 : 7 : 42,$ then the first of these terms in the expansion is