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Logical Reasoning
Letter Series
1!+2!+3!+....+n!=?
Read Solution (Total 4)
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- n!+(n-1)!+ . . . +3!+2!+1!=n![1+(1/n)+(1/(n(n-1)))+ . . . +(2!/n!)+(1!/n!]
Neglect the Higher terms...
=n!{1+(1/n)+(1/(n(n-1))}
=n*n!/(n-1) - 11 years agoHelpfull: Yes(15) No(6)
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- 11 years agoHelpfull: Yes(1) No(2)
- go with option it will more sufficient
last digit would be 13 - 11 years agoHelpfull: Yes(0) No(0)
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- 11 years agoHelpfull: Yes(0) No(0)
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