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What is the value of
1(1!) + 2(2!) + 3(3!) + 4(4!) + ............+ 2012(2012)! ?
Read Solution (Total 4)
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- suppose we take 1(1!) + 2(2!) , then the result is 5 which is 3! - 1.
In the same way if we take 1(1!) + 2(2!) + 3(3!) , then the result is 23 which is 4! - 1.
so for the above series the ans will be 2013! - 1. - 11 years agoHelpfull: Yes(6) No(0)
- (1+2+3....2012)+(1!+2!+3!...2012!)
=(2012*(2012+1)/2)+(2013!-1) - 11 years agoHelpfull: Yes(2) No(2)
- 2011(2011)!
- 7 years agoHelpfull: Yes(0) No(0)
- (n+1)!-1
so (2013)!-1 - 7 years agoHelpfull: Yes(0) No(0)
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