How do I prove this by induction?

Question

thank you for taking the time to help me with the question. I am struggling to use proof by induction for this formula:

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

So far, I came up with:

$$S(n) = (n + 1)! – 1$$

Base case: When $n = 0$,

$(n + 1)! - 1 = (1)! - 1 = 0 = S(0)$, thus the base case is proved.

Inductive step: Suppose S$(n) = (n + 1)! – 1$, and n > 1 Then we have... $$S(n + 1) = ((n + 1) + 1)! - 1$$

I don't understand what is required of the proof by induction. What am I supposed to do?

Answer 1

Using your function $S$ seems confusing to me - and I think it might be confusing you too. There are two functions at play here and we want to show that they are equal. In particular, let's define: $$S_1(n)=\sum_{k=0}^nk\times k!$$ $$S_2(n)=(n+1)!-1.$$ The goal is to prove that these two functions are equal. Your current proof looks like it's trying to show $S_2(n)=S_2(n)$ - which is probably why it's not working out so well.

So, you can start with the base case where $S_1(0)=S_2(0)=0$, which can you can by computation. Then, the inductive step is as follows

Suppose that $S_1(n)=S_2(n)$. Prove that $S_1(n+1)=S_2(n+1)$.

You can do this by relating the value of $S_1(n+1)$ to the value of $S_1(n)$ and the value of $S_2(n+1)$ to $S_2(n)$.