(1!*1)+(2!*2)+(3!*3).......(n!*n)the sum to n terms of the above series is?

• (1!*1)+(2!*2)+(3!*3).......(n!*n =(n+1)! -1
  
  check, when
  n=2,
  (1!*1)+(2!*2)= 1+4=5 = 3!-1
  when n=3,
  (1!*1)+(2!*2)+(3!*3)= 1+4+18 = 23=4!-1
  
• 12 years agoHelpfull: Yes(2) No(0)
• x = (1!*1)+(2!*2)+(3!*3).......(n!*n)
  y = 1! + 2! + 3!.......... n!
  x + y = (1!*2)+(2!*3)+(3!*4).......(n!*n+1)
  x + y = 2! + 3! + ...........(n+1)!
  x + y + 1 = 1! + 2! + 3! +......(n+1)!
  x + y + 1 = y + (n + 1)!
  x + 1 = (n + 1)!
  x = (n + 1)! - 1
  enjoyyy......
• 12 years agoHelpfull: Yes(1) No(1)
• 0≤ k ≤n
  ==>∑k!*k
  =∑[k!*{(k+1)-1}]
  =∑[k!*(k+1)-k!]
  =∑(k+1)!- ∑k!
  =(n+1)!-1
• 12 years agoHelpfull: Yes(0) No(0)