find the sum of the series given below 1(1!)+2(2!)+3(3!)+.........+2012(2012!)

• let x = 1(1!)+2(2!)+3(3!)+.........+2012(2012!) & y = 1!+2!+3!+.........+2012!
  x+y = 2(1!)+3(2!)+....+2013(2012!) = 2!+3!+....+2012!+2013!
  x+y+1 = 1!+2!+3!+.........+2012!+2013! = y+2013!
  x = 2013! -1
  (solved)
• 12 years agoHelpfull: Yes(44) No(1)
• (N+1)!-1
  hence, ans is 2013!-1
• 12 years agoHelpfull: Yes(8) No(4)
• 
  1*1! + 2*2! + 3*3! +...+2012*2012!
  Now
  S (1) = 1*1! = 1
  S (2) = 1*1! + 2*2! = 1+4 = 5 = 3! - 1
  S (3) = 1+4+18 = 23 = 4! - 1
  .
  .
  S (n) = (n+1)! - 1
  So sum of this series is
  S (2012) = 2013! - 1
• 8 years agoHelpfull: Yes(8) No(0)
• x=1(1!)+2(2!)+......+2012(2012!) & y= 1!+2!+3!+....+2012!
  x+y= 1!(1+1)+2!(2+1)+.......+2011!(2011+1)+2012!(2012+1)
  = 2(1!)+3(2!)+....+2012(2011!)+2013(2012!)
  =2!+3!+.....+2012!+2013!
  x+y+1=1!+2!+....+2012!+2013! =y+2013!
  x=y+2013!-y-1
  x=2013!-1
• 9 years agoHelpfull: Yes(3) No(0)
• how you solved it?plz explain
  
• 12 years agoHelpfull: Yes(2) No(2)
• let x = 1(1!)+2(2!)+3(3!)+.........+2012(2012!) & y = 1!+2!+3!+.........+2012!
  x+y = 2(1!)+2(2!)+....+2013(2012!) = 2!+3!+....+2012!+2013!
  x+y+1 = 1!+2!+3!+.........+2012!+2013! = y+2013!
  x = 2013! -1
  (solved)
  
• 12 years agoHelpfull: Yes(2) No(3)
• i not get it
• 12 years agoHelpfull: Yes(0) No(0)
• 2012 is factorial is 4046132
  and 2012!*2012 is 8140817584
• 11 years agoHelpfull: Yes(0) No(4)
• i can't understand please explain clearly
• 8 years agoHelpfull: Yes(0) No(0)
• (n+1)!-1
  =2013!-1
• 6 years agoHelpfull: Yes(0) No(0)