### Related Formula sum_k=1^infty left( frac1k! - frac1(k+3)! right) implies textTelescoping Series simplification$\sum_{k=1}^{\infty} \left( \frac{1}{k!} - \frac{1}{(k+3)!} \right) \implies \text{Telescoping Series simplification}$ ### Core Logic Rewrite the numerator polynomial to establish factor terms matching the factorial expansion base (k+3)$(k+3)$: k^3 + 6k^2 + 11k + 5 = (k^3 + 6k^2 + 11k + 6) - 1 = (k+1)(k+2)(k+3) - 1$k^3 + 6k^2 + 11k + 5 = (k^3 + 6k^2 + 11k + 6) - 1 = (k+1)(k+2)(k+3) - 1$ ### Step 1: Simplify General Term T_k = frac(k+1)(k+2)(k+3)(k+3)! - frac1(k+3)!$T_k = \frac{(k+1)(k+2)(k+3)}{(k+3)!} - \frac{1}{(k+3)!}$ T_k = frac1k! - frac1(k+3)!$T_k = \frac{1}{k!} - \frac{1}{(k+3)!}$ This creates a clean telescoping layout format structure. ### Step 2: Sum the Series Writing out expanded \partial sums up to infinity: S = left( frac11! + frac12! + frac13! + frac14! + dots right) - left( frac14! + frac15! + frac16! + dots right)$S = \left( \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots \right) - \left( \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \dots \right)$ All higher terms cancel out systematically, leaving exactly the leading remaining fragments: S = frac11! + frac12! + frac13! = 1 + frac12 + frac16 = frac106 = frac53$S = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} = 1 + \frac{1}{2} + \frac{1}{6} = \frac{10}{6} = \frac{5}{3}$ ### Pattern Recognition Whenever factorials dominate fraction denominators, manipulate structural terms to align components via Telescoping sums (V_n - V_n-k$V_n - V_{n-k}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limits, Continuity and Differentiability Class 11 Mathematics: Sequences and Series