Question

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

Answer 1

Answer with explanation:

The given sequaence whose sum we have to find

`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}}`

→k*(k+1)=(k+1)!