Sequences & Series
Geometric series
Grade 11
Question:
<p>In a geometric series, the first term is <em>a</em> and common ratio is <em>r</em>. If \(S_n\) denotes the sum of the <em>n</em> terms and \(U_n = \sum_{n=1}^{n} S_n\), then \(rS_n + (1-r)U_n\) equals</p>
<p>(1) 0</p>
<p>(2) \(n\)</p>
<p>(3) \(na\)</p>
<p>(4) \(nar\)</p>
Step-by-Step Solution
Key Concept: Express both S_n and U_n in terms of a and r, then recognize that U_n is a telescoping sum of partial sums. The linear combination rS_n + (1-r)U_n will simplify to a form independent of n.
<p><strong>Step 1:</strong> Write S_n for geometric series: S_n = a(1-r^n)/(1-r) [when r ≠ 1]</p><p><strong>Step 2:</strong> Find U_n = Σ(k=1 to n) S_k = Σ(k=1 to n) [a(1-r^k)/(1-r)]</p><p>U_n = [a/(1-r)] · Σ(k=1 to n)(1-r^k) = [a/(1-r)] · [n - (r + r^2 + ... + r^n)]</p><p>U_n = [a/(1-r)] · [n - r(1-r^n)/(1-r)]</p><p><strong>Step 3:</strong> Compute rS_n + (1-r)U_n</p><p>rS_n = ra(1-r^n)/(1-r)</p><p>(1-r)U_n = a[n - r(1-r^n)/(1-r)] = an - ar(1-r^n)/(1-r)</p><p><strong>Step 4:</strong> Add them:</p><p>rS_n + (1-r)U_n = ra(1-r^n)/(1-r) + an - ar(1-r^n)/(1-r) = an</p><p>∴ Answer: <strong>an</strong></p>
Correct Answer: C