Question

`1+2*2!+3*3!+ ... + n*(n!)=`?

Answer 1

`(n+1)! - 1`

Explanation:

`sum_(r=1)^n r* r! = sum_(r=1) ^n(r+1)r! - r!`

Recall : `(n+1)n! = (n+1)!`

`=> sum_(r=1) ^n (r+1)! - r!`

Method of differences:

`color(red)(r = 1) : cancel(2!) color(blue)(-1!`

`color(red)(r = 2) : cancel(3!) - cancel(2!)`
`.`
`.`
`.`
`color(red)(r = n-1) : cancel(n!) - cancel((n-1)!)`
`color(red)( r= n) : color(blue)((n+1)!) - cancel(n!)`

`color(blue)(=>(n+1)! - 1`

`=> sum_(r=1)^n r *r! = (n+1)! - 1`

Answer 2

`sum_(k=1)^n k * k! = (n+1)! - 1`

Explanation:

One way of solving this problem is to spot a pattern, then prove it by induction.

Looking at the first few sums, we find:

`1 * 1! = 1`

`1 * 1! + 2 * 2! = 1+4 = 5`

`1 * 1! + 2 * 2! + 3 * 3! = 1+4+18 = 23`

Note that we should expect a sum that involves a factorial somewhere and `23 = 24-1 = 4!-1`.

Then looking at the previous values we have `5 = 6-1 = 3!-1` and `1 = 2-1 = 2!-1`

So it looks like `sum_(k=1)^n k * k! = (n+1)! - 1`

Let's try to prove it by induction:

Let `P(n)` be the proposition that `sum_(k=1)^n k * k! = (n+1)! - 1`

Base case

Note that:

`sum_(k=1)^color(blue)(1) k * k! = 1 * 1! = 1 = 2-1 = 2!-1 = (color(blue)(1)+1)! - 1`

So `P(1)` is true.

Induction step

Suppose `P(n)` is true for some `n`.

Then:

`sum_(k=1)^(color(blue)(n+1)) k * k! = sum_(k=1)^n k * k! + (n+1)(n+1)!`

`color(white)(sum_(k=1)^(n+1) k * k!) = (n+1)! - 1 + (n+1)(n+1)!`

`color(white)(sum_(k=1)^(n+1) k * k!) = (n+2)(n+1)! - 1`

`color(white)(sum_(k=1)^(n+1) k * k!) = ((color(blue)(n+1))+1)! - 1`

So `P(n+1)` is true.

Conclusion

Having shown `P(1)` and `P(n) => P(n+1)` we can conclude that `P(n)` holds for all integers `n >= 1`.