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Find the value of 1*1! + 2*2! + 3*3! + ......+n*n!
(1) n! +1
(2) (n+1)!
(3) (n+1)!-1
(4) (n+1)!+1
(1) n! +1
(2) (n+1)!
(3) (n+1)!-1
(4) (n+1)!+1
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07 May 2005, 09:38
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christoph
I guess you mean 1*1! + 2*2!....
if only 1 term, 1*1! = 1 ==> only (3) fits, where n in the choises is the number of terms..
if 2 terms, 1*1! + 2*2! = 5 ==> (3)
so it is (3) by backsolving.
08 May 2005, 10:22
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(3) it is...
just pick number upto 3 and see which ones add up!
just pick number upto 3 and see which ones add up!
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Thank you for understanding, and happy exploring!
