UNSW MathSoc Championship Q6

In general, the polynomial can be written as $a(z+1)(z+2) \cdots (z+10)$ . Since $a_0 = 1$ , we know that $a_0 z^{10 - 0} = z^{10}$ is the highest-order term, which forces $a = 1$ . Then finding the value of $a_2$ means finding a term in the form $a_2 z^{10 - 2}$ , which is just the coefficient of $z^8$ .

A coefficient of $z^8$ arises when we multiply by $z$ $8$ times, and a constant the other $2$ times. If we write this out plainly, it is hard to calculate: $S = 1 \cdot 2 + 1 \cdot 3 + 1 \cdot 4 + \cdots + 2 \cdot 3 + 2 \cdot 4 + 2 \cdot 5 + \cdots + 3 \cdot 4 + 3 \cdot 5 + 3 \cdot 6 + \cdots$ . However, if we allow each term to be written twice, for example where a term appears once as $1 \cdot 2$ , and another time in the form $2 \cdot 1$ , and add the terms $1 \cdot 1 + 2 \cdot 2 + \cdots + 10 \cdot 10$ , then we can easily compute the sum.

Following this process gives us $1 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 + \cdots + 2 \cdot 1 + 2 \cdot 2 + 2 \cdot 3 + \cdots + 3 \cdot 1 + 3 \cdot 2 + 3 \cdot 3 + \cdots$ , which equals $1 \cdot (1 + 2 + 3 + \cdots + 10) + 2 \cdot (1 + 2 + 3 + \cdots + 10) + \cdots$ or $1 \cdot 55 + 2 \cdot 55 + \cdots 10 \cdot 55 = 55^2 = 3025$ , where the $55$ comes from the sum of consecutive numbers from 1 to 10: $\frac{10(10+1)}{2}$ . We then remove the terms $1^2 + 2^2 + \cdots + 10^2$ since we cannot choose to multiply terms in the same bracket twice, and the sum of these terms is $\frac{10(10+1)(2\cdot 10+1)}{6} = 385$ .

Therefore, $2S = 3025 - 385$ since we have duplicated each term twice, so $S = \frac{3025 - 385}{2} = \boxed{1320}$ .

If $P(z) = \Pi_{j=1}^{10} (z-j)$ , then the coefficient for the $z^8$ term can be calculated by Vieta's Formulae according to:

$a_{2} = \Sigma_{p=1}^{9} \Sigma_{q=p+1}^{10} p \cdot q = \boxed{1,320}.$