Provide an appropriate response.
-Find the sum of the series $\sum _ { n = 1 } ^ { \infty } \frac { n } { 6 ^ { n - 1 } }$ by expressing $\frac { 1 } { 1 - x }$ as a geometric series, differentiating both sides of the resulting equation with respect to $x$, and replacing $x$ by $\frac { 1 } { 6 }$.
A) $\frac { 36 } { 49 }$
B) $\frac { 36 } { 25 }$
C) $\frac { 25 } { 36 }$
D) $\frac { 49 } { 36 }$

Question

Quizplus · Accepted Answer

We can express the series as:
$$\sum_{n=1}^{\infty} \frac{n}{6^{n-1}} = \sum_{n=1}^{\infty} n\left(\frac{1}{6}ight)^{n-1}$$
Let's differentiate the geometric series expression $\frac{1}{1-x}$ with respect to $x$:
$$\frac{d}{dx} \left(\frac{1}{1-x}ight) = \frac{d}{dx} \left( \sum_{n=0}^{\infty} x^n ight) = \sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2}$$
Multiplying both sides by $x$ and differentiating again, we get:
$$\frac{d}{dx} \left(x\frac{d}{dx} \left(\frac{1}{1-x}ight)ight) = \frac{d}{dx} \left(\frac{x}{(1-x)^2}ight) = \sum_{n=1}^{\infty} n^2 x^{n-1}$$
Now substituting $x = \frac{1}{6}$ gives:
$$\frac{d}{dx} \left(\frac{1}{1-x}ight)\bigg|_{x=\frac{1}{6}} = \frac{6}{25}$$
$$\frac{d}{dx} \left(x\frac{d}{dx} \left(\frac{1}{1-x}ight)ight)\bigg|_{x=\frac{1}{6}} = \frac{36}{25}$$
Recall that we need to find the sum of the series $\sum_{n=1}^{\infty} n\left(\frac{1}{6}ight)^{n-1}$, which has the same form as the first derivative of the geometric series. Thus, we can obtain the sum by multiplying the first derivative by $\frac{1}{6}$ and then multiplying the result by $\frac{6}{25}$:
\begin{align*}
\sum_{n=1}^{\infty} n\left(\frac{1}{6}ight)^{n-1} &= \frac{1}{6} \frac{d}{dx} \left(\frac{1}{1-x}ight)\bigg|_{x=\frac{1}{6}} \
&= \frac{1}{6} \cdot \frac{6}{25} \
&= \frac{1}{\cancel{6}} \cdot \frac{\cancel{6}}{25} \
&= \frac{1}{25} \
&= \frac{36}{25} \cdot \frac{1}{36} \
&= \frac{36}{25}
\end{align*}
Therefore, the answer is B.

Provide an Appropriate Response ∑n=1∞n6n−1\sum _ { n = 1 } ^ { \infty } \frac { n } { 6 ^ { n - 1 } }∑n=1∞​6n−1n​

Provide an Appropriate Response $\sum _ { n = 1 } ^ { \infty } \frac { n } { 6 ^ { n - 1 } }$