Find the sum of the series (1*1!+2*2!+3*3!+...........+2012*2012!)

• General term is n*n!=((n+1)-1)*(n!)
  =(n+1)*n!-1*n!
  =(n+1)!-n!
  S=(2!-1!)+(3!-2!)+(4!-3!)+.........+(2013!-2012!)
  S=2013!-1!
  =2013!-1
• 9 years agoHelpfull: Yes(29) No(2)
• 1*1!=1=2!-1
  1*1!+2*2!=5=3!-1
  1*1!+2*2!+3*3!=23=4!-1
  and so on....
  1*1!+2*2!+3*3!+...........+2012*2012!=2013!-1
  
• 9 years agoHelpfull: Yes(17) No(0)
• n*n!
  = ((n+1)-1)*n!
  =(n+1)*n!- n!
  =(n+1)!-n!
  
  putting values of n=1,2,......,2012
  (2!-1!)+(3!-2!)+(4!-3!)+.........+(2013!-2012!)
  =2013!-1
• 9 years agoHelpfull: Yes(4) No(0)
• (1*1!+2*2!+3*3!+...........+2012*2012!)
  it can be written as
  (2-1)1!+(3-1)2!+(4-1)3!+.........(2013-1)2012!
  2!-1!+3!-2!+4!-3!+............2013!-2012!
  2013!-1 ans
• 9 years agoHelpfull: Yes(1) No(0)
• 2013!-1 as it satisfies the pattern
• 9 years agoHelpfull: Yes(0) No(0)
• add n and subtract n in each term where n=1,2,3,....2012. we get 2013!-1!.
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• sum of series =2013!-1 [n*n!=(n+1)!-n!]
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• 2013!-1;
  E(n.n!)=E(n+1-1)n!=E(n+1)!-E(n!);
  implies, ans is 2013!-1;
• 9 years agoHelpfull: Yes(0) No(0)