$C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \dots + C_{n-r} C_n =$

If $^{2017}C_0 + ^{2017}C_1 + ^{2017}C_2 + ...... + ^{2017}C_{1008} = \lambda^2$ where $\lambda > 0$,then the remainder when $\lambda$ is divided by $33$ is:

If $C_j = {}^{n}C_j$,then $C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n = $

If $(\frac{1}{^{15}C_{0}}+\frac{1}{^{15}C_{1}})(\frac{1}{^{15}C_{1}}+\frac{1}{^{15}C_{2}})...(\frac{1}{^{15}C_{12}}+\frac{1}{^{15}C_{13}}) = \frac{a^{13}}{^{14}C_{0} \cdot ^{14}C_{1} \cdot ... \cdot ^{14}C_{12}}$, then $30a$ is equal to:

$\sum_{r=1}^{15} r^2 \left( \frac{{}^{15}C_r}{{}^{15}C_{r-1}} \right) = $

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