1! + 2! + ��. + 50!=?

• I think there is no such shortcut but if you provide options then it can be solve easily by just finding unit digit...
  
  For example in above question:
  
  1! = 1 (units digit 1)
  2! = 2 (units digit 2)
  3! = 6 (units digit 6)
  4! = 24 (units digit 4)
  5!, 6!.... the units digit will be 0 because all the terms will include 5x2=10
  
  units digit of 1!+2!+... will be 1+2+6+4+0+0+0.... = 3
  
  Hence the answer should contain unit digit as 3.
  
  
  
  Or you can try(I m not sure)
  Use (n+1)!=(n+1)⋅n!=n⋅n!+n!.
• 10 years agoHelpfull: Yes(29) No(1)
• There are no such shortcuts for finding the summation of a factorial series. But if you know the factorial of the last number it helps.
  50! is 30414093201713378043612608166064768844377641568960512000000000000
  (Of course I used calculator for it)
  
  now rearranging the series as
  50!+49!+48!+.....+1!
  = 50!(1+1/50+1/(50*49)+1/(50*48*49)+1/(50*48*49*47).....)(ignoring the rest due to extremely small values.
  =50!(1+0.02+0.00040816326+0.0000085034+...)
  =50!*1.020409
  =3.1034814e+64
  
  Which is not 100% accurate, but is 99.9% accurate.
• 10 years agoHelpfull: Yes(6) No(0)
• 1!+2!+3!+...............................50!
  =50!*(1!/50! + 2!/50! + ........................49!/50!+1)
  =50!*(1+49!/(50*49!) + 48!/(50*49*48!) + ........................................+1!/50!)
  =50!*(1+1/50+1/(50*49)+...................................................)
  =50!*(1+0.02) (Taking the 1st 2 terms rest of the terms are neglegible)
  =1.02*50!
• 10 years agoHelpfull: Yes(1) No(0)
• SUM=l/2(a+l)
  =50!/2(1!+50!)
  =50!*51!/2
• 10 years agoHelpfull: Yes(0) No(3)