A) \[3\,(n\,!)\,+\,n-3\]
B) \[(n+1)!\,-\,(n-1)!\]
C) \[(n+1)\,!\,-1\]
D) \[2\,(n\,!)-2n-1\]
Correct Answer: C
Solution :
\[{{S}_{n}}=1(1!)+2(2!)+3(3!)+.....+n(n!)\] =\[(2-1)(1!)+(3-1)(2!)+(4-1)(3!)+.....\]\[+[(n+1)-1](n!)\] = \[(2.1!-1!)+(3.2!-2!)+(4.3!-3!)+....\]\[+[(n+1)(n!)-(n!)]\] =\[(2!-1!)+(3!-2!)+(4!-3!)+....+[(n+1)!-(n)!]\] = \[(n+1)!-1!\].You need to login to perform this action.
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