If the sum of the coefficients of $x^7$ and $x^{14}$ in the expansion of $\left(\frac{1}{x^3} - x^4\right)^n, x \neq 0$, is zero, then the value of $n$ is . . . . . .

Let $s_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$s_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $s_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
Statement $-1$: $s_3 = 55 \times 2^9$
Statement $-2$: $s_1 = 90 \times 2^8$ and $s_2 = 10 \times 2^8$

In the expansion of $(x-1)(x-2) \ldots (x-18)$, the coefficient of $x^{17}$ is:

The coefficient of $x^7$ in the expansion of $(1 - x - x^2 + x^3)^6$ is

The $r^{th}$ term in the expansion of $(a + 2x)^n$ is

The coefficient of $x^{n-6}$ in the expansion $n! \left[ x - \left( \frac{^nC_0 + ^nC_1}{^nC_0} \right) \right] \left[ \frac{x}{2} - \left( \frac{^nC_1 + ^nC_2}{^nC_1} \right) \right] \left[ \frac{x}{3} - \left( \frac{^nC_2 + ^nC_3}{^nC_2} \right) \right] \dots \left[ \frac{x}{n} - \left( \frac{^nC_{n-1} + ^nC_n}{^nC_{n-1}} \right) \right]$ is equal to: