In order not to complicate, I'll not state the full problem statement, just the part that needs to be proven. We should prove that $x< [1-(\frac{2}{3})^n]$ , for natural $n > 0$.
In the solution, they state that the following is true:
$2*[1-(\frac{2}{3})^{n-1}]+1 = 3*[1-(\frac{2}{3})^{n}]$
My guess is that this is the ceiling function,.
The square brackets are standard parentheses. The equality can be proved via the following series of operations: $$ 2\cdot\left[1 - \left(\frac{2}{3}\right)^{n-1}\right] + 1 = 2 + 1 - 2\left(\frac{2}{3}\right)^{n-1} = 3 - \frac{2^n}{3^{n-1}} = 3 - 3\frac{2^n}{3^{n}} = 3\left[1 - \left(\frac{2}{3}\right)^{n}\right] $$