Find the sum of $\sum_{n = 0}^{\infty} \frac{n (n + 1)}{3^{n - 1}}$ by identifying it as the value of the derivative of the series
$g (x) = \sum_{n = 0}^{\infty} (n + 1) x^{n} = \frac{1}{(1 - x)^{2}} for | x | < 1 .$
a. Take the derivative of the series given by $g (x)$ .
Write the derivative of the series in the first box and the derivative of the rational expression in the second box.
$g^{'} (x) = \sum_{n = 1}^{\infty} ◻ =$
b. Find the value of $g^{'} (\frac{1}{3}) = \sum_{n = 0}^{\infty} \frac{n (n + 1)}{3^{n - 1}}$ .
$g^{'} (\frac{1}{3}) =$

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Finally, we find that the sum of the series $\sum_{n = 0}^{\infty} \frac{n (n + 1)}{3^{n - 1}}$ is $6.75$ .

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Step 1 ：First, we need to find the derivative of the series $g (x)$ . The derivative of the series $g (x)$ is $g^{'} (x) = \frac{2}{(1 - x)^{3}}$ .

Step 2 ：Next, we substitute $x = \frac{1}{3}$ into the derivative to find the sum of the series. Substituting $x = \frac{1}{3}$ into the derivative, we get $g^{'} (\frac{1}{3}) = 6.75$ .

Step 3 ：Finally, we find that the sum of the series $\sum_{n = 0}^{\infty} \frac{n (n + 1)}{3^{n - 1}}$ is $6.75$ .

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