Progression (Sequence and Series)HardBloom L3
Question
The sum $\sum_{k = 1}^{n}\mspace{2mu} k!\left( k^{2} + k + 1 \right)$ is equal to
Options
A.$(n + 1)(n + 1)! - 1$
B.$nn! - 1$
C.$(n + 1)! - 1$
D.$n(n + 1)! - 1$
Solution
Sol. [Key Concept: $\frac{k}{(k^{2 +}k + 1)\ ?}$ — write the term as a difference so the factorial sum telescopes.]
$$\ \sum_{k = 1}^{n}\mspace{2mu} k!\left( (k + 1)^{2} - k \right) = \sum_{k = 1}^{n}\mspace{2mu}((k + 1)(k + 1)! - k(k!))$$
$$= (n + 1)(n + 1)! - 1$$
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