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Question

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

[AMU 1999; DCE 2005]


A
3(n!) + n - 3
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B
(n + 1)! - (n - 1)!
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C
(n + 1)! - 1
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D
2(n!) - 2n - 1
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Solution

The correct option is C (n + 1)! - 1
Sn = 1(1!) + 2(2!) + 3(3!) + ...... + n(n!)
= (2 - 1)(1!) + (3 - 1)(2!) + (4 - 1)(3!) + .... +[(n + 1) -1] (n!)
= (2.1! - 1!) + (3.2! - 2!) + (4.3! - 3!) + .... +[(n + 1)(n!) -(n!)]
= (2! - 1!) + (3! - 2!) + (4! - 3!) + ....+ [(n + 1)! - (n)!]
= (n + 1)! - 1!.

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