$\left(\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \frac{1}{10 \cdot 13} + \frac{1}{13 \cdot 16}\right)$ is equal to

### Concept & Formula This series is a variation of the telescoping sum where the factors in the denominator have a constant difference $d$. We can split the terms using partial fraction decomposition. $$\frac{1}{a \cdot b} = \frac{1}{b - a} \left( \frac{1}{a} - \frac{1}{b} \right)$$ ### Step-by-Step Solution Identify the constant difference $d$ between the factors in each denominator: $4 - 1 = 3$ $7 - 4 = 3$ $10 - 7 = 3$ So, the common difference $d = 3$. Apply the partial fraction formula to each term: $$\frac{1}{1 \cdot 4} = \frac{1}{3} \left( \frac{1}{1} - \frac{1}{4} \right)$$ $$\frac{1}{4 \cdot 7} = \frac{1}{3} \left( \frac{1}{4} - \frac{1}{7} \right)$$ $$\vdots$$ $$\frac{1}{13 \cdot 16} = \frac{1}{3} \left( \frac{1}{13} - \frac{1}{16} \right)$$ Add the terms together and factor out $\frac{1}{3}$: $$\text{Sum} = \frac{1}{3} \left[ \left(\frac{1}{1} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{7}\right) + \cdots + \left(\frac{1}{13} - \frac{1}{16}\right) \right]$$ The intermediate terms cancel out perfectly. $$\text{Sum} = \frac{1}{3} \left[ 1 - \frac{1}{16} \right]$$ $$= \frac{1}{3} \left[ \frac{15}{16} \right]$$ $$= \frac{5}{16}$$ ### Exam Strategy & Shortcut Use the general shortcut for constant-difference telescoping series: $$\text{Sum} = \frac{1}{\text{Difference}} \times \left( \frac{1}{\text{First Factor}} - \frac{1}{\text{Last Factor}} \right)$$ Here, Difference = $3$, First Factor = $1$, Last Factor = $16$. $$\text{Sum} = \frac{1}{3} \times \left( 1 - \frac{1}{16} \right) = \frac{1}{3} \times \frac{15}{16} = \frac{5}{16}$$ This bypasses writing out the full sequence. ### Common Pitfall The most frequent mistake is forgetting to divide by the difference $d = 3$. Many students simply compute $(1 - \frac{1}{16})$ and arrive at $\frac{15}{16}$, completely missing the scaling factor. ### Final Answer **Therefore, the correct answer is 5/16.**