How to verify by induction that 1(1!) + 2(2!) + ... + n(n!) = (n+1)! - 1 for every pos. int. n?

Question

Basis step: $n=1: 1(1!) = (1+1)! - 1 = 1$, true;

$n=2 : 1(1!) + 2(2!) = 5 = (2+1)! - 1 = 6 - 1$, true;
$n=3 : 1(1!) + 2(2!) + 3(3!) = 23 = (3+1)! - 1 = 24 - 1$, true;
...

How do I prove the statement as a whole using mathematical induction?

Answer 1

Hint: $(k+1)!-1+(k+1)(k+1)! = (k+1)!(k+2) - 1 = ?$

Answer 2

To prove a statement $P(n)$ by mathematical induction, you must first prove the base step $P(1)$, which allows you to assume that $P(m)$ holds for some positive integer $m$ (since it holds for $n = 1$), then establish that $P(m) \Rightarrow P(m + 1)$. This establishes that $P(n)$ holds for all positive integers $n$ since $P(1) \Rightarrow P(2) \Rightarrow P(3) \Rightarrow \cdots$.

Let $P(n)$ be the statement that $$\sum_{k = 1}^n k \cdot k! = (k + 1)! - 1$$

Let $n = 1$. Then

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

so $P(1)$ holds (as you established above).

Assume $P(m)$ holds for some positive integer $m$. Then

$$\sum_{k = 1}^m k \cdot k! = (m + 1)! - 1$$

Let $n = m + 1$. Then

\begin{align*} \sum_{k = 1}^{m + 1} k \cdot k! & = \sum_{k = 1}^m k \cdot k! + (m + 1)(m + 1)!\\ & = (m + 1)! - 1 + (m + 1)(m + 1)! && \text{by the induction hypothesis}\\ & = (1 + m + 1)(m + 1)! - 1\\ & = (m + 2)(m + 1)! - 1\\ & = (m + 2)! - 1\\ & = [(m + 1) + 1]! - 1 \end{align*} so $P(m) \Rightarrow P(m + 1)$. Thus, $P(n)$ holds for each positive integer $n$.