Understanding an inductive proof that $ \sum_{i=1}^n i\times i! = (n +1)!-1$

Question

I can't get my head around understanding this proof:

Problem: Prove that $ \sum_{i=1}^n i\times i! = (n +1)!-1$, by induction.

Solution:

Base case: $\sum_{i=1}^1 i\times i! = 1 = (1+1)!-1 = 2 - 1 = 1$

Assume: $\sum_{i=1}^n i\times i! = (n+1)! - 1$

Solution:

$$\sum_{i=1}^{n+1} i\times i! = (n+1) \times (n+1)! + \sum_{i=1}^ni \times i!$$ (here the largest term is pulled from the sum) $$\sum_{i=1}^{n+1}i\times i! = (n+1)\times (n+1)! + (n+1)!-1 $$(substituted the sum with our assumption) $$=(n+1)!\times [(n+1)+1]-1$$ $$=(n+2)!-1$$

The last two lines are where I can't find how they are done. I know for one it is easy, but any help understanding it and recommendations for resources to read are welcomed. Regards.

Answer 1

First, the author factors out $(n+1)!\;$ in the two first terms: $$(n+1)\times\color{red}{(n+1)!}+\color{red}{(n+1)!}=\bigl((n+1)+1\bigr)\color{red}{(n+1)!}=(n+2)\times\color{red}{(n+1)!}$$ then use the recursive definition of factorials: $\;(n+2)\times (n+1)!=(n+2)!$.

Answer 2

In the inductive step assuming that

$$\sum_{i=1}^n i\times i! = (n +1)!-1$$

is true, we have just to prove that

$$\sum_{i=1}^{n+1}i\times i! = (n+2)!-1$$

then you are done, indeed

$$...=(n+1)!\times ((n+1)+1)-1=(n+2)(n+1)!-1=(n+2)!-1$$

Answer 3

Alternatively: $$E=\sum_{i=1}^n (i+1-1)\cdot i!=\sum_{i=1}^n ((i+1)!-i!)=$$ $$2!-1!+3!-2!+\cdots +(n+1)!-n!=(n+1)!-1.$$