If $(1+x)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n$ and $a_0 - a_2 + a_4 - a_6 + \ldots = k \cos \frac{n \pi}{4}$,then $k = $

$\sum_{r=0}^{10} {}^{40-r} C_5$ is equal to

Let $S_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $S_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
Assertion $(A) : S_3 = 55 \times 2^9$
Reason $(R) : S_1 = 90 \times 2^8$ and $S_2 = 10 \times 2^8$

If the coefficients of $x^9, x^{10}$ and $x^{11}$ in the expansion of $(1+x)^n$ are in arithmetic progression,then $n^2-41n$ is equal to

The sum of the coefficients of the last $19$ terms in the binomial expansion of $(1+x)^{37}$ is

The value of $\frac{C_1}{C_0} + 2 \cdot \frac{C_2}{C_1} + 3 \cdot \frac{C_3}{C_2} + \dots + n \cdot \frac{C_n}{C_{n-1}}$ is equal to