......................find the sum of 1!+2!+3!+4!+...100!

• 1!+2!+3!+4!+...100!
  =100!(1+1/100+1/99*100+...+1/100!)
  =100!(1+.01+.000101+....)
  =100!*1.01(approax)
  =100!*101/100
  101!/100
• 11 years agoHelpfull: Yes(33) No(2)
• its simply (n+1)!/n=101!/100
  
• 11 years agoHelpfull: Yes(10) No(6)
• (n+1)!/n is not valid for n=4
• 11 years agoHelpfull: Yes(5) No(2)
• i don't think (n+1)!/n is evolves.for example,sum of 1!+2!+3!=9...but by implementing the suspicious formula it gives value 8.... so far it gives us the right one? so please don't give up.... try to sort out i'm unable to do it :(
• 11 years agoHelpfull: Yes(4) No(1)
• no ..(n+1)!/n will not satisfy all the values of n...:(
• 11 years agoHelpfull: Yes(2) No(2)
• (N+1)!/n=101!/100
• 11 years agoHelpfull: Yes(0) No(5)
• (n+1)!/n i.e: 101!/100
• 11 years agoHelpfull: Yes(0) No(4)
• sum of 1!+2!+...+ n!=(n+1)!/n
  since n=100
  1!+2!+3!+...100!=101!/100
• 11 years agoHelpfull: Yes(0) No(3)
• (n+1)!/100
• 11 years agoHelpfull: Yes(0) No(4)
• plz anyone solve the sum of 1!+2!+.....50!
• 10 years agoHelpfull: Yes(0) No(0)