## Task A-4.1.

Let $a_{0}, a_{1}, \ldots, a_{n}$ be real numbers such that

$$
a_{0}+a_{1} x+\ldots+a_{n} x^{n}=(x+1)^{3}(x+2)^{3} \cdots(x+672)^{3}
$$

Determine the sum

$$
a_{2}+a_{4}+a_{6}+\ldots+a_{2016}
$$
## Solution.

Let $P(x)=a_{0}+a_{1} x+\ldots+a_{n} x^{n}=(x+1)^{3}(x+2)^{3} \cdots(x+672)^{3}$. This is a polynomial of degree $3 \cdot 672=2016$.

Notice that

$$
\begin{aligned}
P(1) & =a_{0}+a_{1}+a_{2}+a_{3}+\ldots+a_{2014}+a_{2015}+a_{2016} \\
P(-1) & =a_{0}-a_{1}+a_{2}-a_{3}+\ldots+a_{2014}-a_{2015}+a_{2016}
\end{aligned}
$$

By adding these, we get

$$
P(1)+P(-1)=2 a_{0}+2 a_{2}+\ldots+2 a_{2014}+2 a_{2016}
$$

Since $a_{0}=P(0)$, we have

$$
a_{2}+a_{4}+\ldots+a_{2016}=\frac{1}{2}(P(1)+P(-1))-P(0)
$$

We calculate:

$$
\begin{aligned}
P(0) & =1^{3} \cdot 2^{3} \cdots 672^{3}=(672!)^{3}, & & 1 \text{ point} \\
P(1) & =2^{3} \cdot 3^{3} \cdots 673^{3}=(673!)^{3}, & & 1 \text{ point} \\
P(-1) & =0 & & 1 \text{ point}
\end{aligned}
$$

Finally, we get

$$
a_{2}+a_{4}+\ldots+a_{2016}=\frac{1}{2}(P(1)+P(-1))-P(0)=\frac{1}{2}(673!)^{3}-(672!)^{3}
$$

1 point