If $\alpha \, and \, \beta$ are the roots of the equation $6x^2 \, - \, 6x \, + \, 1 \, = \,0$ then $1/2 \left ( p \, + \, q\alpha \, + \, r\alpha ^2 \, + \, s\alpha ^3 \right ) \, + \, 1/2\left ( p \, + \, q\beta \, + \, r\beta ^2 \, + \,s\beta ^3 \right )$ is $\dfrac{p}{1} \, + \,\dfrac{q}{2} \, + \, \dfrac{r}{3} \, + \, \dfrac{s}{4}.$

Factoring out the 1/2 from the expression gives 1/2 * ((p + p) + (q * alpha + q * beta) + (r * alpha^2 + r * beta^2) + (s * alpha^3 + s * beta^3)). Using Vieta's formulas for 6x^2 - 6x + 1 = 0, the sum of the roots is alpha + beta = 1, the product is alpha * beta = 1/6, and alpha^2 + beta^2 = 1 - 1/3 = 2/3. For the cubic term, alpha^3 + beta^3 = (alpha + beta)^3 - 3 * alpha * beta * (alpha + beta) = 1 - 1/2 = 1/2. Substituting these values into the grouped expression gives 1/2 * (2p + q(1) + r(2/3) + s(1/2)), which simplifies to p + q/2 + r/3 + s/4, making the statement True.

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