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

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(C) Let the sum be $S_n = 1(1!) + 2(2!) + 3(3!) + \dots + n(n!)$.We can rewrite the $k$-th term as $k(k!) = ((k + 1) - 1)k! = (k + 1)! - k!$.Now,we su

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$(n + 1)! - 1$

Vedclass · Answer

(C) Let the sum be $S_n = 1(1!) + 2(2!) + 3(3!) + \dots + n(n!)$.We can rewrite the $k$-th term as $k(k!) = ((k + 1) - 1)k! = (k + 1)! - k!$.Now,we sum these terms from $k = 1$ to $n$:$S_n = \sum_{k=1}^{n} ((k + 1)! - k!)$$S_n = (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + ((n + 1)! - n!)$This is a telescoping series where intermediate terms cancel out.$S_n = (n + 1)! - 1!$$S_n = (n + 1)! - 1$.

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

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