$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \frac{1}{4 \cdot 5 \cdot 6}$ is equal to

### Concept & Formula This is an advanced telescoping series where each denominator has three consecutive factors. We can split each term into the difference of two fractions, each containing two factors. The formula to decompose a term with three factors is: $$\frac{1}{k(k+1)(k+2)} = \frac{1}{2} \left[ \frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)} \right]$$ ### Step-by-Step Solution Apply the decomposition formula to each term in the series: $$\frac{1}{1 \cdot 2 \cdot 3} = \frac{1}{2} \left[ \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} \right]$$ $$\frac{1}{2 \cdot 3 \cdot 4} = \frac{1}{2} \left[ \frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} \right]$$ $$\frac{1}{3 \cdot 4 \cdot 5} = \frac{1}{2} \left[ \frac{1}{3 \cdot 4} - \frac{1}{4 \cdot 5} \right]$$ $$\frac{1}{4 \cdot 5 \cdot 6} = \frac{1}{2} \left[ \frac{1}{4 \cdot 5} - \frac{1}{5 \cdot 6} \right]$$ Now, add all these terms together and factor out the common $\frac{1}{2}$: $$\text{Sum} = \frac{1}{2} \left[ \left(\frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3}\right) + \left(\frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4}\right) + \dots + \left(\frac{1}{4 \cdot 5} - \frac{1}{5 \cdot 6}\right) \right]$$ Notice the cascade of cancellations. Every intermediate term cancels out, leaving only the first component of the first term and the last component of the final term: $$\text{Sum} = \frac{1}{2} \left[ \frac{1}{1 \cdot 2} - \frac{1}{5 \cdot 6} \right]$$ $$= \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{30} \right]$$ Find a common denominator to subtract the fractions inside the bracket: $$= \frac{1}{2} \left[ \frac{15}{30} - \frac{1}{30} \right]$$ $$= \frac{1}{2} \left[ \frac{14}{30} \right] = \frac{7}{30}$$ ### Exam Strategy & Shortcut For a series of $n$ terms in the format $\sum \frac{1}{k(k+1)(k+2)}$, the generalized sum formula is: $$\frac{n(n+3)}{4(n+1)(n+2)}$$ Here, we have exactly $4$ terms, so $n = 4$: $$S_4 = \frac{4(4+3)}{4(4+1)(4+2)} = \frac{4 \times 7}{4 \times 5 \times 6} = \frac{28}{120} = \frac{7}{30}$$ Memorizing this generalized formula saves immense time during competitive exams. ### Common Pitfall A common error is attempting to compute the actual value of each fraction ($\frac{1}{6} + \frac{1}{24} + \frac{1}{60} + \frac{1}{120}$) and finding the LCM. While doable for 4 terms, this method becomes impossible under time pressure if the series has 10 or 20 terms. ### Final Answer **Therefore, the correct answer is 7/30.**