<p>Find the sum to \(n\) terms of the series,<br>\[1 + \left(1 + \dfrac{1}{2} + \dfrac{1}{2^2}\right) + \left(1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \dfrac{1}{2^4}\right) + \cdots\]</p>
Step-by-Step Solution
Key Concept: Recognize that the k-th group contains (2k-1) terms forming a geometric series, then sum the individual group sums using the formula for geometric series and telescoping.
<p><strong>Step 1: Identify the pattern of groups</strong></p><p>Group 1: 1 term = 2(1)-1</p><p>Group 2: 3 terms = 2(2)-1</p><p>Group 3: 5 terms = 2(3)-1</p><p>The k-th group has (2k-1) terms of a geometric series with first term 1 and common ratio 1/2.</p><p><strong>Step 2: Find the sum of the k-th group</strong></p><p>Sum of k-th group = $\dfrac{1 - (1/2)^{2k-1}}{1 - 1/2} = 2\left(1 - \dfrac{1}{2^{2k-1}}\right)$</p><p><strong>Step 3: Sum all n groups</strong></p><p>$S_n = \sum_{k=1}^{n} 2\left(1 - \dfrac{1}{2^{2k-1}}\right) = 2n - 2\sum_{k=1}^{n} \dfrac{1}{2^{2k-1}}$</p><p><strong>Step 4: Evaluate the geometric sum</strong></p><p>$\sum_{k=1}^{n} \dfrac{1}{2^{2k-1}} = \dfrac{1}{2} + \dfrac{1}{2^3} + \dfrac{1}{2^5} + \cdots = \dfrac{1/2(1-(1/4)^n)}{1-1/4} = \dfrac{2}{3}\left(1 - \dfrac{1}{2^{2n}}\right)$</p><p><strong>Step 5: Final answer</strong></p><p>∴ $S_n = 2n - 2 \cdot \dfrac{2}{3}\left(1 - \dfrac{1}{2^{2n}}\right) = 2n - \dfrac{4}{3}\left(1 - \dfrac{1}{2^{2n}}\right)$</p>
Correct Answer: \(2n - \dfrac{4}{3}\left(1 - \dfrac{1}{2^{2n}}\right)\)