<s>[INST] 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.
 [/INST] By hypothesis, we know that
$$
\frac{1}{3}+\frac{1}{3^2}+\dots+\frac{1}{3^k}=\frac{1}{2}\bigg[1-\frac{1}{3^k}\bigg]
$$
Hence, by adding to both sides by $\frac{1}{3^{k+1}}$ we get
$$\begin{align}
\frac{1}{3}+\frac{1}{3^2}+\dots+\frac{1}{3^k}+\frac{1}{3^{k+1}}&=\frac{1}{2}\bigg[1-\frac{1}{3^k}\bigg]+\frac{1}{3^{k+1}}\\
&=\frac{3^k-1}{2\cdot 3^k}+\frac{1}{3^{k+1}}\\
&=\frac{3^{k+1}-3+2}{2\cdot 3^{k+1}}\\
&=\frac{3^{k+1}-1}{2\cdot 3^{k+1}}\\
&=\frac{1}{2}\bigg[1-\frac{1}{3^{k+1}}\bigg]
\end{align}$$
</s>