Let $p(x) = 1 + x + x^2 + x^3 + oldots + x^{12}.$ What is the coefficient of $x^8$ in the expansion of $p(x + 1)$? (A) 1 (B) 495 (C) 715 (D) 1287 - HSC - SSCE Mathematics Extension 2 - Question 6 - 2016 - Paper 1

Now, substituting $x + 1$ into $p(x)$:

$p(x + 1) = \frac{1 - (x + 1)^{13}}{1 - (x + 1)} = \frac{1 - (x + 1)^{13}}{-x}$ This can be rewritten as:

$-\frac{1 - (x + 1)^{13}}{x}$ To find the coefficient of $x^8$, we need to expand $(x + 1)^{13}$ using the binomial theorem.

The binomial theorem states that:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ In our case, we set $a = x$ and $b = 1$:

$(x + 1)^{13} = \sum_{k=0}^{13} \binom{13}{k} x^{13-k} 1^k = \sum_{k=0}^{13} \binom{13}{k} x^{13-k}$

To find the coefficient of $x^8$, we need the term where $13 - k = 8$, or $k = 5$. Thus, the required coefficient is:

$\binom{13}{5}$

The binomial coefficient is calculated as:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$ Substituting $n = 13$ and $k = 5$:

$\binom{13}{5} = \frac{13!}{5! (13-5)!} = \frac{13!}{5! 8!}$ Calculating this yields:

$\binom{13}{5} = \frac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1} = 1287$

Thus, the coefficient of $x^8$ in the expansion of $p(x + 1)$ is:

1287

Therefore, the correct answer is (D) 1287.