Smallest Set of Weights for Two-Pan Balance/Mistake

Source Work

 * The Dictionary
 * $31$
 * $31$

Mistake

 * Using both pans, the solution is similar, but now relies on expressing the weight as the sum and difference of powers of $3$. With the weights $1$, $3$, $9$ and $27$ it is possible to weight up to $40$. In general the weights up to $3$ will weigh up to a maximum of $\frac 1 2 \left({3^{n + 1} - 1}\right)$.

In that last sentence there is a typo.

It should say:
 * ... the weights up to $3^n$ will weigh ...