Question

Use induction to prove that for all integers n 2 1 we have 1.1! +2.2! +3.3! + ... +nin! = (n + 1)! - 1.

Answer 1

Let
`P(n)` be the statement that

`1\cdot1!+2\cdot2!+\cdots+n\cdot n!=(n+1)!-1`

For
`n=1`, we have
`P(1)` true, since

`1\cdot1!=1`

and

`(1+1)!-1=2!-1=2-1=1`

Assume
`P(k)` is true. Then

`1\cdot1!+2\cdot2!+\cdots+k\cdot k!=(k+1)!-1`

Then for
`n=k+1` we have

`1\cdot1!+2\cdot2!+\cdots+k\cdot k!+(k+1)\cdot(k+1)!=(k+1)!-1+(k+1)\cdot(k+1)!`

because
`P(k)` is assumed to be true, then

`(k+1)!-1+(k+1)(k+1)!=(k+1)!(1+(k+1))-1=(k+2)\cdot(k+1)!-1=(k+2)!-1`

and so
`P(n)` is true for all
`n`. QED

Alternatively, no induction is needed, since

`(n+1)!=(n+1)\cdot n!=n\cdot n!+n!`

so

`1\cdot1!+2\cdot2!+\cdots+n\cdot n!=(2!-1!)+(3!-2!)+\cdots+((n+1)!-n!)=(n+1)!-1!`

`\implies1\cdot1!+2\cdot2!+\cdots+n\cdot n!=(n+1)!-1`