jmc

We have that $$\sum _{i=0}^{50}{a}_i{x}^i=\left(1-\frac{1}{3}x+\frac{1}{6}{x}^2\right)\left(1-\frac{1}{3}{x}^3+\frac{1}{6}{x}^6\right)\cdots \left(1-\frac{1}{3}{x}^9+\frac{1}{6}{x}^{18}\right).$$If we multiply this out (which we're not going to do), this involves taking a term from the first factor $1-\frac{1}{3}x+\frac{1}{6}{x}^2,$ a term from the second factor $1-\frac{1}{3}{x}^3+\frac{1}{6}{x}^6,$ and so on, until we take a term from the fifth factor $1-\frac{1}{3}{x}^9+\frac{1}{6}{x}^{18},$ and taking the product of these terms.

Suppose the product of the terms is of the form $c{x}^n,$ where $n$ is even. Then the number of terms of odd degree, like $-\frac{1}{3}x$ and $-\frac{1}{3}{x}^3,$ that contributed must have been even. These are the only terms from each factor that are negative, so $c$ must be positive.

Similarly, if $n$ is odd, then the number of terms of odd degree that contributed must be odd. Therefore, $c$ is negative. Hence, $$\begin{array}{rl}\sum _{i=0}^{50}|a_i|& =|a_0|+|a_1|+|a_2|+\cdots +|a_{50}|\\ & =a_0-a_1+a_2-\cdots +a_{50}\\ & =Q(-1)\\ & =P(-1{)}^5\\ & ={\left(1+\frac{1}{3}+\frac{1}{6}\right)}^5\\ & =\boxed{\frac{243}{32}}.\end{array}$$