Sequences & Series
Sum to n Terms of Special Series
Grade 11
Question:
<p>The sum of the first <i>n</i> terms of the series \(1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \cdots\) is \(\dfrac{n(n+1)^2}{2}\) when <i>n</i> is even. When <i>n</i> is odd the sum is</p>
<p>\(\dfrac{3n(n+1)}{2}\)</p>
<p>\(\dfrac{n^2(n+1)}{2}\)</p>
<p>\(\dfrac{n(n+1)^2}{4}\)</p>
<p>\(\left[\dfrac{n(n+1)}{2}\right]^2\)</p>
Step-by-Step Solution
Key Concept: Identify the pattern alternates between single and double coefficients: separate odd-positioned and even-positioned terms, then use the even case formula to derive the odd case by recognizing the structural relationship.
<p><strong>Step 1: Identify the pattern</strong></p><p>The series is: 1² + 2·2² + 3² + 2·4² + 5² + 2·6² + ...</p><p>The pattern alternates: single odd-indexed squares, then double even-indexed squares.</p><p><strong>Step 2: Recognize the structure</strong></p><p>When n is even: S_n = n(n+1)²/2</p><p>When n is odd (say n = 2k+1), we have k complete pairs plus one additional odd-indexed term (which is (2k+1)²).</p><p><strong>Step 3: Relate odd to even case</strong></p><p>If n is odd, then S_n = S_(n-1) + (n)²</p><p>Since (n-1) is even: S_(n-1) = (n-1)n²/2</p><p><strong>Step 4: Calculate for odd n</strong></p><p>S_n = (n-1)n²/2 + n²</p><p>S_n = (n-1)n²/2 + 2n²/2</p><p>S_n = [n²(n-1) + 2n²]/2</p><p>S_n = [n³ - n² + 2n²]/2</p><p>S_n = [n³ + n²]/2</p><p>S_n = n²(n+1)/2</p><p>∴ Answer: <strong>n²(n+1)/2</strong> (or equivalently n(n+1)²/2 when expressed differently)</p>
Correct Answer: B