Sequences & Series
Summation of Series / Telescoping
Grade 11
Question:
<p>If \(S = \displaystyle\sum_{n=1}^{5} \frac{1}{n(n+1)(n+2)(n+3)} = \frac{k}{3}\), then \(k\) equals:</p>
<p>\(\dfrac{55}{336}\)</p>
<p>\(\dfrac{55}{112}\)</p>
<p>\(\dfrac{1}{336}\)</p>
<p>\(\dfrac{55}{100}\)</p>
Step-by-Step Solution
Key Concept: Use partial fractions decomposition: 1/[n(n+1)(n+2)(n+3)] = 1/3[1/(n(n+1)(n+2)) - 1/((n+1)(n+2)(n+3))]. This creates a telescoping series where consecutive terms cancel, leaving only the first and last terms.
<p><strong>Step 1: Set up partial fractions</strong></p><p>Use the identity: <br>1/[n(n+1)(n+2)(n+3)] = (1/3)[1/(n(n+1)(n+2)) - 1/((n+1)(n+2)(n+3))]</p><p><strong>Step 2: Recognize telescoping nature</strong></p><p>When we sum from n=1 to 5:</p><p>S = (1/3)[1/(1·2·3) - 1/(2·3·4)] + (1/3)[1/(2·3·4) - 1/(3·4·5)] + ... + (1/3)[1/(5·6·7) - 1/(6·7·8)]</p><p><strong>Step 3: Apply telescoping cancellation</strong></p><p>All middle terms cancel. Only the first and last terms survive:</p><p>S = (1/3)[1/(1·2·3) - 1/(6·7·8)]</p><p>S = (1/3)[1/6 - 1/336]</p><p><strong>Step 4: Simplify</strong></p><p>1/6 - 1/336 = 56/336 - 1/336 = 55/336 = 55/336</p><p>S = (1/3) · (55/336) = 55/1008 = 5/96</p><p>Since S = k/3, we have k/3 = 5/96</p><p>∴ <strong>k = 5/32</strong> (or verify: 5/96 × 3 gives the relationship)</p><p><strong>Rechecking: k = 5/32</strong></p>
Correct Answer: A