1(1!)+2(2!)+3(3!)+----------+2012(2012!)

• 1(1!)=1
  1(1!)+2(2!)=5
  1(1!)+2(2!)+3(3!)=23
  1(1!)+2(2!)+3(3!)+4(4!)=119
  1(1!)+2(2!)+3(3!)+4(4!)+5(5!)=719
  So 1(1!)+2(2!)+3(3!)+--------+n(n!)={(n+1)!}-1
  Hence 1(1!)+2(2!)+3(3!)+-------+2012(2012!)=(2012+1)!-1=(2013!)-1
• 12 years agoHelpfull: Yes(73) No(6)
• summation of [sigma] n*n! where n is any positive integer number from 1 to n
• 12 years agoHelpfull: Yes(0) No(2)
• 0 because 2012-2012=0
  
• 12 years agoHelpfull: Yes(0) No(21)