<p>Find the sum to \(n\) terms: \[1 + \left(1 + \frac{1}{2} + \frac{1}{2^2}\right) + \left(1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3}\right) + \ldots\]</p>
Step-by-Step Solution
Key Concept: Recognize that the rth term is a geometric series with sum 2 - 1/2^(r-1), then sum these results using telescoping or direct summation of the resulting arithmetic-geometric series.
<p><strong>Step 1:</strong> Identify the rth term as a geometric series: </p><p>Term_r = 1 + 1/2 + 1/2² + ... + 1/2^(r-1) (r terms)</p><p>This is a finite G.P. with first term a = 1, common ratio r = 1/2, and r terms.</p><p><strong>Step 2:</strong> Apply the G.P. sum formula:</p><p>Term_r = (1 - (1/2)^r)/(1 - 1/2) = 2(1 - 1/2^r) = 2 - 2/2^r = 2 - 1/2^(r-1)</p><p><strong>Step 3:</strong> Sum all n terms:</p><p>S_n = Σ(r=1 to n) (2 - 1/2^(r-1))</p><p>= 2n - Σ(r=1 to n) 1/2^(r-1)</p><p>= 2n - (1 + 1/2 + 1/2² + ... + 1/2^(n-1))</p><p><strong>Step 4:</strong> The second sum is a G.P. with sum = (1 - (1/2)^n)/(1/2) = 2(1 - 1/2^n)</p><p>S_n = 2n - 2(1 - 1/2^n) = 2n - 2 + 2/2^n</p><p>∴ <strong>Answer: 2n - 2 + 1/2^(n-1)</strong> or equivalently <strong>2(n-1) + 1/2^(n-1)</strong></p>
Correct Answer: 2