If $(1-x+x^2)^n = a_0 + a_1 x + \ldots + a_{2n} x^{2n}$, then the value of $a_0 + a_2 + a_4 + \ldots + a_{2n}$ is

The sum of the last eight coefficients in the expansion of $(1 + x)^{15}$ is

Let $\binom{n}{k}$ denote ${}^{n}C_{k}$ and $\left[\begin{array}{c} n \\ k \end{array}\right]=\begin{cases} \binom{n}{k}, & \text{if } 0 \leq k \leq n \\ 0, & \text{otherwise} \end{cases}$. If $A_{k}=\sum_{i=0}^{9}\binom{9}{i}\left[\begin{array}{c} 12 \\ 12-k+i \end{array}\right]+\sum_{i=0}^{8}\binom{8}{i}\left[\begin{array}{c} 13 \\ 13-k+i \end{array}\right]$ and $A_{4}-A_{3}=190p$,then $p$ is equal to:

Find the arithmetic mean of $^nC_0, ^nC_1, ^nC_2, \dots, ^nC_n$.

If the coefficients of $x^4, x^5$ and $x^6$ in the expansion of $(1+x)^n$ are in arithmetic progression,then the maximum value of $n$ is:

If ${ }^{n} C_0+\frac{1}{2}{ }^{n} C_1+\frac{1}{3}{ }^{n} C_2+\ldots+\frac{1}{n+1}{ }^{n} C_{n}=\frac{1023}{10}$,then $n=$