The sum 1(1!) + 2(2!) + 3(3!) + dots + n(n!) | vedclass

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(C) Let the sum be $S_n = \sum_{k=1}^{n} k(k!)$.We can rewrite the general term $k(k!)$ as $(k + 1 - 1)k!$.$k(k!) = (k + 1)k! - k! = (k + 1)! - k!$.No

Vedclass · Accepted Answer

$(n + 1)! - 1$

Vedclass · Answer

(C) Let the sum be $S_n = \sum_{k=1}^{n} k(k!)$.We can rewrite the general term $k(k!)$ as $(k + 1 - 1)k!$.$k(k!) = (k + 1)k! - k! = (k + 1)! - k!$.Now,substitute this into the summation:$S_n = \sum_{k=1}^{n} ((k + 1)! - k!)$.This is a telescoping series:$S_n = (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + ((n + 1)! - n!)$.All intermediate terms cancel out,leaving:$S_n = (n + 1)! - 1! = (n + 1)! - 1$.

The sum $1(1!) + 2(2!) + 3(3!) + \dots + n(n!)$ equals

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