<p>Find the sum of \(n\) terms of the series \(\dfrac{4}{3} + \dfrac{10}{9} + \dfrac{28}{27} + \cdots\)</p>
Step-by-Step Solution
Key Concept: Decompose each term as a difference of two simpler fractions, then separate into two telescoping/standard series. Recognize the general term pattern by examining numerators (4, 10, 28) which follow 2(3^k - 1) + 2 = 2·3^k.
<p><strong>Step 1:</strong> Find the general term. Examining the series:</p><ul><li>Term 1: 4/3 = (6-2)/3</li><li>Term 2: 10/9 = (18-8)/9</li><li>Term 3: 28/27 = (54-26)/27</li></ul><p>The denominator is 3^n. For numerators: observe pattern → <strong>General term = [2·3^n - 2]/3^n = 2 - 2/3^n</strong></p><p><strong>Step 2:</strong> Split and sum separately:</p><p>$$S_n = \sum_{k=1}^{n}\left(2 - \frac{2}{3^k}\right) = \sum_{k=1}^{n}2 - 2\sum_{k=1}^{n}\frac{1}{3^k}$$</p><p><strong>Step 3:</strong> Evaluate each part:</p><ul><li>First part: $\sum_{k=1}^{n}2 = 2n$</li><li>Second part: $\sum_{k=1}^{n}\frac{1}{3^k} = \frac{1/3(1-(1/3)^n)}{1-1/3} = \frac{1}{2}\left(1 - \frac{1}{3^n}\right)$</li></ul><p><strong>Step 4:</strong> Combine:</p><p>$$S_n = 2n - 2 \cdot \frac{1}{2}\left(1 - \frac{1}{3^n}\right) = 2n - 1 + \frac{1}{3^n}$$</p><p>$$= \frac{3^n(2n-1) + 1}{3^n} = \frac{3^n(2n+1) - 1}{2 \times 3^n}$$</p><p>∴ Answer: $\dfrac{3^n(2n+1)-1}{2 \times 3^n}$</p>
Correct Answer: \(\dfrac{3^n(2n+1)-1}{2 \times 3^n}\)