2!+3!+4!+5!+.....+100!

• ans is (101!/100)-1
• 11 years agoHelpfull: Yes(13) No(1)
• 100!+99!+98!+...=100!(1+1/100+1/(100*99)+......too small so we ignore it)=
  100!(1+.01+.00001+....)=100!*1.01 (approx) am i right???
• 11 years agoHelpfull: Yes(11) No(4)
• unit place will be equal to sum of unit place of 2!+3!+4!(5! to 100! contains 0 at unit place)
  =2+6+24=32
  so unit place will be 2
  
  101!+2
  
• 11 years agoHelpfull: Yes(3) No(0)
• =2+3*2+4*3*2+5*4*3*2+..... 100*99*98*...2
  = 2*(1+3*(1+4*(1+5*(.....*(1+100)
  =101*(100)*(99)*(98)*(97)....*2*1
  =101!
• 11 years agoHelpfull: Yes(2) No(5)
• ans is 101!/100
• 11 years agoHelpfull: Yes(1) No(0)
• 100(101!)-1
• 11 years agoHelpfull: Yes(0) No(10)
• try to take 2!+3!+4! for the first time and generalize it for as many times as required and eliminate the choices.
• 11 years agoHelpfull: Yes(0) No(0)
• 1!+2!+...100! -1!
  101!/100 -1!
  101(100!)/100 -1 ans
• 11 years agoHelpfull: Yes(0) No(0)
• 99*100!+0
  hence proved
• 11 years agoHelpfull: Yes(0) No(0)
• ans should be 101!/3
• 11 years agoHelpfull: Yes(0) No(3)