Step 1: Understanding the Concept:
This problem involves the sum of infinite Geometric Progressions (GPs). We split the summation into two parts and sum each part individually. : Key Formula or Approach:
Sum of infinite GP \( S_{\infty} = \frac{a}{1-r} \) for \( |r|<1 \). Step 2: Detailed Explanation:
The expression is \( S = \sum_{r=1}^{\infty} [3 \cdot (\frac{1}{2})^{r} - 2 \cdot 3 \cdot (\frac{1}{3})^{r}] \).
Note: \( 3^{1-r} = 3^1 \cdot 3^{-r} = 3 \cdot (\frac{1}{3})^r \).
So, \( S = 3 \sum_{r=1}^{\infty} (\frac{1}{2})^{r} - 6 \sum_{r=1}^{\infty} (\frac{1}{3})^{r} \).
1. Evaluate first part: \( \sum_{r=1}^{\infty} (\frac{1}{2})^{r} = \frac{1/2}{1 - 1/2} = \frac{1/2}{1/2} = 1 \).
2. Evaluate second part: \( \sum_{r=1}^{\infty} (\frac{1}{3})^{r} = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2} \).
Combine:
\[ S = 3(1) - 6(1/2) = 3 - 3 = 0 \]. Step 3: Final Answer:
The value of the summation is 0.