vandarughb747 vandarughb747
  • 07-08-2019
  • Mathematics
contestada

If 5 t n2, then 5t m (5 points) (5 points) 5. 1(11) +2(2!)+-...+nn!)(n+1)!-1 for all n e I

Respuesta :

Аноним Аноним
  • 12-08-2019

Answer with explanation:

The given sequaence whose sum we have to find

  [tex]S_{n}=1*1!+2*2!+3*3!+4*4!+...........+n*n!\\\\T_{n}{\text{general term}}=n*n!\\\\ \sum _{k=1}^{k=n}t_{k}\\\\\sum _{k=1}^{k=n}k(k!)\\\\\sum _{k=1}^{k=n}(k+1-1)(k!)\\\\\sum _{k=1}^{k=n}[(k+1)!-k!]\\\\=(1!+2!+3!+.............(n+1)!)-(1!+2!+3!+..........+n!)\\\\=(n+1)!\\\\----{\text{Cancelling out like terms}}[/tex]

→k*(k+1)=(k+1)!

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