mathematical induction with exponent $$\frac 13 + \frac 1{3^2} + \frac 1{3^3} + \dots + \frac 1{3^n} +  = \frac 12  \times \left( 1 - \frac{1}{3^n} \right)$$
Step 1 - $n=1$
$$\begin{align}
\frac 1 {3^1} & = \frac 1 2 \times \left( 1 - \frac 1 {3^1}   \right) \\
\frac 1 3     & = \frac 1 2 \times \left( 1 - \frac 1 3       \right) \\
\frac 1 3     & = \frac 1 2 \times \frac 2 3                          \\
\frac 1 3     & = \frac 1 3                                           \\
\end{align}$$
Step 2 - n$=k$
$$ \frac 13 + \frac 1{3^2} + \frac1 {3^3} + \dots + \frac 1 {3^k} = \frac 12 \times \left( 1 - \frac 1 {3^k} \right)$$
Step 3 - $n=k+1$
Having problems solving step 3.

You must prove that $$\sum_1^k\frac{1}{3^i} + \frac{1}{3^{k+1}} = \frac{1}{2}\left(1-\frac{1}{3^{k+1}}\right).$$
Applying the assumption that $$\sum_1^k\frac{1}{3^i} = \frac{1}{2}\left(1-\frac{1}{3^k}\right)$$, you can get the result.
