The remainder when $428^{2024}$ is divided by 21 is

If the second, third and fourth terms in the expansion of $(x+y)^n$ are 135,30 and $\frac{10}{3}$, respectively, then $6\left(n^3+x^2+y\right)$ is equal to $\_\_\_\_$ .

Let a $$ \begin{aligned} t a=1+\frac{{ }^2 C_2}{3!}+\frac{{ }^3 C_2}{4!}+\frac{{ }^4 C_2}{5!}+\cdots, & \\ & \quad b=1+\frac{{ }^1 C_0+{ }^1 C_1}{1!}+\frac{{ }^2 C_0+{ }^2 C_1+{...

Let $\alpha=\sum_{\mathrm{r}=0}^{\mathrm{n}}\left(4 \mathrm{r}^2+2 \mathrm{r}+1\right)^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$ and $\beta=\left(\sum_{r=0}^n \frac{{ }^n C_r}{r+1}\right)+\frac{1}{n+1}$. If $140<\frac{2 \alpha}{\beta}<281$, then the value of $n$ is

If the constant term in the expansion of (1+2x-3x^3 ) (3/2 x^2-1/3x)^9 is p, then 108p is equal to

The coefficient of $x^{70}$ in $x^2(1+x)^{98}+x^3(1+x)^{97}+\mathrm{x}^4(1+\mathrm{x})^{96}++x^{54}(1+x)^{46}$ is ${ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}$. Then a possible value to p+q is :

The sum of all rational terms in the expansion of $\left(2^{\frac{1}{5}}+5^{\frac{1}{3}}\right)^{15}$ is equal to :

In the expansion of $(1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5, x \neq 0$, the sum of the coefficient of $x^3$ and $x^{-13}$ is equal to

If the Coefficient of $x^{30}$ in the expansion of $\left(1+\frac{1}{x}\right)^6\left(1+x^2\right)^7\left(1-x^3\right)^8 ; x \neq 0$ is $\alpha$, then $|\alpha|$ equals