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$\left(\forall n \in \mathbb{N}\right)\left((n + 1)! = (n + 1) \cdot n!\right)$

Prove the following statement by induction: for all $n \in \mathbb{N}$

$\sum_{k=0}^{n}(k \cdot k!) = (n + 1)! − 1$

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Base case: $n =0$

$(0\cdot 0!) = 0 ~\wedge~ (0+1)! – 1 = 0$, true.

Assume $n$ is true so $k = n+1$

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

I’m not sure where to carry on from here? Can anyone shed some light?

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$$\sum_{k=0}^{n+1} k{\cdot}k!=\sum_{k=0}^nk{\cdot}k!+(n{+}1){\cdot}(n{+}1)!=(n{+}1)!{-}1+(n{+}1){\cdot}(n{+}1)!=(n{+}2){\cdot}(n{+}1)!{-}1=(n{+}2)!{-}1$$

@ASoni Yes,

(n+1)!−1+(n+1)⋅(n+1)!=(n+1)!+(n+1)⋅(n+1)!−1=(n+1)!(1+n+1)−1(n+1)!−1+(n+1)⋅(n+1)!=(n+1)!+(n+1)⋅(n+1)!−1=(n+1)!(1+n+1)−1

– user2345215 Nov 11 ’14 at 22:56

How do you get from `(n+1)!+(n+1)⋅(n+1)!−1`

to `(n+1)!⋅(1+n+1)−1`

?

Factoring out `(n+1)!`

from `(n+1)!+[(n+1)⋅(n+1)!]−1`

you will have `(n+1)!⋅[1+(n+1)]-1`

.

Similarly, you may wish to factor a `2`

out of only the first two of three terms in `(2+4)-1`

. In which case you will have `[2⋅(1+2)]-1`

.

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