Find the sum of the series: 1 cdot 1! + 2 cdo | vedclass

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Question

(A) Let the $k$-th term of the series be $T_k = k \cdot k!$. We can rewrite $T_k$ as: $T_k = ((k + 1) - 1) \cdot k! = (k + 1) \cdot k! - k! = (k + 1)!

Vedclass · Accepted Answer

$(n + 1)! - 1$

Vedclass · Answer

(A) Let the $k$-th term of the series be $T_k = k \cdot k!$. We can rewrite $T_k$ as: $T_k = ((k + 1) - 1) \cdot k! = (k + 1) \cdot k! - k! = (k + 1)! - k!$. Now,the sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} ((k + 1)! - k!)$. This is a telescoping series: $S_n = (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + ((n + 1)! - n!)$. All intermediate terms cancel out,leaving: $S_n = (n + 1)! - 1! = (n + 1)! - 1$.

Find the sum of the series: $1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \dots + n \cdot n!$

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