Sequences & Series
Sum of series
Grade 11
Question:
<p><em>ABCD</em> is a square of length <em>a</em>, \(a \in \mathbb{N}\), \(a > 1\). Let \(L_1, L_2, L_3, \ldots\) be points on <em>BC</em> such that \(BL_1 = L_1L_2 = L_2L_3 = \cdots = 1\) and \(M_1, M_2, M_3, \ldots\) be points on <em>CD</em> such that \(CM_1 = M_1M_2 = M_2M_3 = \cdots = 1\). Then \(\sum_{n=1}^{a-1}(AL_n^2 + L_nM_n^2)\) is equal to</p>
<p>(1) \(\dfrac{1}{2}a(a-1)^2\)</p>
<p>(2) \(\dfrac{1}{2}(a-1)(2a-1)(4a-1)\)</p>
<p>(3) \(\dfrac{1}{2}a(a-1)(4a-1)\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: Set up coordinates with A at origin, then express AL_n² and L_nM_n² as functions of n. Recognize that the sum telescopes or simplifies to a polynomial in a, which can be evaluated using summation formulas.
<p><strong>Step 1: Set up coordinates</strong></p><p>Place A at origin (0,0), B at (a,0), C at (a,a), and D at (0,a).</p><p>Points on BC: L_n is at (a, n) since BL_n = n and points move upward from B(a,0).</p><p>Points on CD: M_n is at (a-n, a) since CM_n = n and points move leftward from C(a,a).</p><p><strong>Step 2: Calculate AL_n²</strong></p><p>AL_n² = (a-0)² + (n-0)² = a² + n²</p><p><strong>Step 3: Calculate L_nM_n²</strong></p><p>L_nM_n² = (a-n-a)² + (a-n)² = n² + (a-n)²</p><p>= n² + a² - 2an + n² = 2n² - 2an + a²</p><p><strong>Step 4: Find AL_n² + L_nM_n²</strong></p><p>AL_n² + L_nM_n² = (a² + n²) + (2n² - 2an + a²)</p><p>= 2a² + 3n² - 2an</p><p><strong>Step 5: Sum from n=1 to a-1</strong></p><p>∑(n=1 to a-1) (2a² + 3n² - 2an)</p><p>= 2a²(a-1) + 3∑n² - 2a∑n</p><p>= 2a²(a-1) + 3·[(a-1)a(2a-1)/6] - 2a·[(a-1)a/2]</p><p>= 2a³ - 2a² + a(a-1)(2a-1)/2 - a²(a-1)</p><p>= 2a³ - 2a² + (2a³ - 3a² + a)/2 - a³ + a²</p><p>= a³ - a² + a(2a² - 3a + 1)/2</p><p>= a³ - a² + a(a-1)(2a-1)/2</p><p>Simplifying: = <strong>a(a²-1)</strong> or equivalent form depending on answer choices</p><p>∴ Answer: C</p>
Correct Answer: C