1(1!)+2(2!)+3(3!)....2012(2012!) what is the sum of the series?

• 1(1!)+2(2!)=5 => 3!-1
  1(1!)+2(2!)+3(3!)=23 => 4!-1
  1(1!)+2(2!)+3(3!)+.............+2012(2012!)=2013!-1
• 10 years agoHelpfull: Yes(70) No(0)
• 1!=2!-1
  1(1!)+2(2!)=3!-1
  1(1!)+2(2!)+3(3!)=4!-1
  
  simalarly 2012(2012!) Is 2013!-1
• 10 years agoHelpfull: Yes(26) No(0)
• 1(1!)+2(2!)+3(3!)....2012(2012!)=summation of n.n!
  n.n!=(n+1-1)n!=(n+1)n!-n!=(n+1)!-n!
  by putting values of n=1,2,..................2012
  we get 2013!-1
• 10 years agoHelpfull: Yes(3) No(1)
• ans is 2013!-1
  1(1!)+2(2!)..........+2012(2012!)=2013!-1
• 10 years agoHelpfull: Yes(2) No(0)
• 1(1!)+2(2!)+3(3!)....2012(2012!)
  when this type of prob given... just take a look at its maximum range...
  so it will be [(n+1)!-1]
  Hence it will be as (2012+1)!-1=(2013)!-1
  
  BUT IF IT IS LIKE 1!+2!+3!....(2012!)
  THEN, it will be (n+1)!=2013!
• 10 years agoHelpfull: Yes(2) No(2)
• if the given question is 1(1!)+2(2!)+3(3!)....2012(2012!)
  then answer is 2013!-1
  1(1!)+2(2!)+3(3!).................n(n!)=(n+1)!-1
  
• 10 years agoHelpfull: Yes(0) No(0)
• 2012!=2012*2011*2010........*1
  now for 2012(2012!)=2012*2012*2011*...*1
  =(2013*2012*2011*...*1)-1
  =2013!-1
• 10 years agoHelpfull: Yes(0) No(0)
• simple approach to solve this question is to take first 3 number that is mention in equation
  i.e 1(1!)+2(21)+3(3!)
  now cal calculate this 1(1!)+2(21)+3(3!) will come 23 which is 4!-1 so take a 1 more a highest factorial mention and subtract 1 in this case 2013!-1
• 10 years agoHelpfull: Yes(0) No(1)