1*1!+2*2!+3*3!...2012*2012!? How to solve this type of series?

• 1*1! = 1 = 2!-1
  1*1! + 2*2! = 3!-1
  and so on
  sum = 2013!-1
  
• 11 years agoHelpfull: Yes(43) No(2)
• try to find out the pattern....
  1*1!=1 (2!-1)
  1*1!+2*2!=5 (3!-1)
  1*1!+2*2!+3*3!=23 (4!-1)
  .
  so, answer will be (2013!-1)
• 11 years agoHelpfull: Yes(7) No(0)
• (2-1)*1! + (3-1)*2!...(2013-1)*2012!
  
  (2!+3!+4!+...+2013!)-(1!+2!+3!+···+2012!)
  
  
  2013! + 1! = 2013!-1 answer
• 11 years agoHelpfull: Yes(2) No(0)
• ans is 2013!-1
• 11 years agoHelpfull: Yes(0) No(2)
• ans is 2013!-1
• 11 years agoHelpfull: Yes(0) No(3)
• 2013!-1
  
  is the answer .
• 11 years agoHelpfull: Yes(0) No(0)
• use this formula----
  r*r!=(r+1-1)*r!
  =(r+1)r!-r!
  =(r+1)!-r!
  1*1!=2!-1!
  2*2!=3!-2!
  .
  .
  .
  .
  2012*2012!=2013!-2012!
  add all the term,we get
  
  2013!-1(ans)
  
  
• 11 years agoHelpfull: Yes(0) No(0)