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
${T_{r + 1}} = {\,^{12}}{C_r}{\left( {{x^2}} \right)^{12 - r}}{( - yx)^r} = {( - 1)^r}{\,^{12}}{C_r} \cdot {x^{24 - 2r}}{y^r} = {( - 1)^r}{\,^{12}}{C_r} \cdot {x^{24 - r}} \cdot {y^r}$
The value of $x$ in the expression ${[x + {x^{{{\log }_{10}}}}^{(x)}]^5}$, if the third term in the expansion is $10,00,000$
Prove that $\sum\limits_{r = 0}^n {{3^r}{\,^n}{C_r} = {4^n}} $
If for positive integers $r > 1,n > 2$ the coefficient of the ${(3r)^{th}}$ and ${(r + 2)^{th}}$ powers of $x$ in the expansion of ${(1 + x)^{2n}}$ are equal, then
Coefficient of ${t^{12}}$ in ${\left( {1 + {t^2}} \right)^6}\left( {1 + {t^6}} \right)\left( {1 + {t^{12}}} \right)$ is-
If the coefficients of second, third and fourth term in the expansion of ${(1 + x)^{2n}}$ are in $A.P.$, then $2{n^2} - 9n + 7$ is equal to