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
\[
g(x)=\sum_{n=0}^{\infty}(n+1) x^{n}=\frac{1}{(1-x)^{2}} \text { for }|x|< 1 .
\]
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.
\[
g^{\prime}(x)=\sum_{n=1}^{\infty} \square=
\]
b. Find the value of $g^{\prime}\left(\frac{1}{3}\right)=\sum_{n=0}^{\infty} \frac{n(n+1)}{3^{n-1}}$.
\[
g^{\prime}\left(\frac{1}{3}\right)=
\]
Answer
Finally, we find that the sum of the series \(\sum_{n=0}^{\infty} \frac{n(n+1)}{3^{n-1}}\) is \(\boxed{6.75}\).
Step 1 ：First, we need to find the derivative of the series \(g(x)\). The derivative of the series \(g(x)\) is \(g^\prime(x)=\frac{2}{(1-x)^{3}}\).
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 \(g^\prime\left(\frac{1}{3}\right)=6.75\).
Step 3 ：Finally, we find that the sum of the series \(\sum_{n=0}^{\infty} \frac{n(n+1)}{3^{n-1}}\) is \(\boxed{6.75}\).
