Proving $\sum_{i=1}^n i(i!) = (n+1)! -1$ by Mathematical Induction

Question

Theorem: For any integer n $\ge$ 1.

$$\sum_{i=1}^n i(i!) = (n+1)! -1$$ Prove by mathematical induction.

I have this problem and I know how to go about it, but I don't understand what I should do with the last part of the proof. What I did was:

1. I made the property P(n) equal to $$\sum_{i=1}^n i(i!) = (n+1)! -1$$
2. I then used P(1) as the basis, so I plugged in 1 into n like so: $$\sum_{i=1}^1 i(i!) = (1+1)! -1$$ The left hand side of the equation is equal to 1 and (2)! - 1 is equal to 1 for the right hand side. Thus, since the left hand and the right hand equal to each other, the statement is true.
3. I then assumed P(k) was true. $$\sum_{i=1}^k i(i!) = (k+1)! -1$$
4. Afterwards, I had to prove P(k+1) was true. $$\sum_{i=1}^{k+1} i(i!) = (k+2)! -1$$ So: $$\sum_{i=1}^k (k+1)! - 1 + (k+1)[(k+1)!] = (k+2)! -1$$

I really don't understand step 4 much, but so far am I doing it correct? If so, I don't understand how to make those two equations equal so the proof can be correct. Any help?

Answer 1

Your work is correct up to the point where you attempted to prove $P(k) \implies P(k + 1)$. Let $n = k + 1$. Then \begin{align*} \sum_{i = 1}^{k + 1} i(i!) & = \sum_{i = 1}^{k} i(i!) + (k + 1)(k + 1)!\\ & = (k + 1)! - 1 + (k + 1)(k + 1)! & \text{by the induction hypothesis}\\ & = (1 + k + 1)(k + 1)! - 1\\ & = (k + 2)(k + 1)! - 1\\ & = (k + 2)! - 1 \end{align*} Hence, $P(k) \implies P(k + 1)$. Since $P(1)$ holds and $P(k) \implies P(k + 1)$, $P(n)$ holds for all positive integers.

Answer 2

Hint:

$$\sum_{i=1}^k i(i!)=1(1!)+2(2!)+3(3!)+\cdots+k(k!)$$ while $$\sum_{i=1}^{k+1} i(i!)=1(1!)+2(2!)+3(3!)+\cdots+k(k!)+(k+1)((k+1)!)$$

so if you add a single term, namely $(k+1)((k+1)!))$, to both sides, then the left hand side will be what you need.