<p>If \(\displaystyle\sum_{n=1}^{5} \dfrac{1}{n(n+1)(n+2)(n+3)} = \dfrac{k}{3}\), then \(k\) is equal to</p>
<p>\(\dfrac{55}{336}\)</p>
<p>\(\dfrac{17}{105}\)</p>
<p>\(\dfrac{1}{6}\)</p>
<p>\(\dfrac{19}{112}\)</p>
Step-by-Step Solution
Key Concept: Use partial fraction decomposition to express 1/(n(n+1)(n+2)(n+3)) as a telescoping series where consecutive terms cancel, leaving only boundary terms.
<p><strong>Step 1:</strong> Decompose using the telescoping identity:</p><p>$$\frac{1}{n(n+1)(n+2)(n+3)} = \frac{1}{3}\left[\frac{1}{n(n+1)(n+2)} - \frac{1}{(n+1)(n+2)(n+3)}\right]$$</p><p>This can be verified by finding a common denominator on the right side.</p><p><strong>Step 2:</strong> Apply the sum from n=1 to 5:</p><p>$$\sum_{n=1}^{5} \frac{1}{n(n+1)(n+2)(n+3)} = \frac{1}{3}\sum_{n=1}^{5}\left[\frac{1}{n(n+1)(n+2)} - \frac{1}{(n+1)(n+2)(n+3)}\right]$$</p><p><strong>Step 3:</strong> The series telescopes. Most terms cancel:</p><p>$$= \frac{1}{3}\left[\frac{1}{1·2·3} - \frac{1}{6·7·8}\right]$$</p><p>$$= \frac{1}{3}\left[\frac{1}{6} - \frac{1}{336}\right]$$</p><p><strong>Step 4:</strong> Simplify:</p><p>$$= \frac{1}{3}\left[\frac{56-1}{336}\right] = \frac{1}{3}·\frac{55}{336} = \frac{55}{1008}$$</p><p><strong>Step 5:</strong> Reduce and express as k/3:</p><p>$$\frac{55}{1008} = \frac{55}{1008} = \frac{k}{3}$$</p><p>$$k = \frac{55·3}{1008} = \frac{165}{1008} = \frac{55}{336}$$</p><p>Simplifying: $\gcd(55,336) = 1$, so $k = \frac{55}{336}$ or if the answer expects an integer, verify $k = \frac{55}{336}$ reduces. Testing: $1008 = 336·3$, so $k = \frac{55}{336}$. If answer is integral, $\boxed{k = \frac{55}{336}}$ or context may give $k=55, 336$ etc.</p><p>∴ Answer: A</p>
Correct Answer: A