Equating series of Sums and Products!

We make the following Key observation: if $(x_1, x_2, \ldots, x_{2015})$ satisfies the property:

$\sum_{k=1}^{2015} 2^{k-1}(x_k)^{2015} = 2014\prod_{k=1}^{2015} x_k \quad ....(1)$

then $x_1$ is even.

Substituting $2y_1$ for $x_1$ in (1) and dividing both sides by 2, we get:

$(x_2)^{2015} + 2(x_3)^{2015} + \ldots + 2^{2013}(x_{2015})^{2015} + 2^{2014}(y_1)^{2015}$

$= 2014 \left( \prod_{k=2}^{2015} x_k \right) y_1$

This means that the sequence $(x_2, \ldots, x_{2015}, y_1)$ also satisfies (1).

Iterating, we obtain successively:

$x_2$ is even, $x_2 = 2y_2$ and $(x_3, \ldots, x_{2015}, y_1, y_2)$ satisfies (1).

............

$x_{2015}$ is even, $x_{2015} = 2y_{2015}$ and $(y_1, y_2, \ldots, y_{2015})$ also satisfies (1).

Thus, we have obtained the following result: if $(x_1, x_2, \ldots, x_{2015})$ satisifies (1), then the $x_k$ 's are even and $\left( \frac{x_1}2, \frac{x_2}2, \ldots, \frac{x_{2015}}2 \right)$ also satisfies (1).

It follows that all $x_k$ 's are $0$ . Indeed, if $x_j \neq 0$ (say), then we might write $x_j = 2^r \cdot z, r \geq 1, z$ being odd; iterating the previous process $r$ times would yield a sequence satisfying (1) and containing the odd term $z$ , which is impossible, as we have seen. Conversely, the null sequence obviously satisfies (1) so that we may conclude that the only sequence satisfying (1) is the null sequence.

Therefore, the Sum is obviously $\boxed{0}$ .