Prove $\sum_{i=1}^n i! \cdot i = (n+1)! - 1$?

Question

Prove the summation: $$\sum_{i=1}^n i! \cdot i = (n+1)! - 1$$ using induction.

base case: $n=1$: \begin{align*} \sum_{i=1}^1 i! \cdot i &= (1+1)! - 1 \\ 1 &= 2 - 1 \\ 1 &= 1 \end{align*}

This is a question from my test review packet, currently have the base case completed and I am a bit lost on where to go from there. Any help/hints are appreciated.

Answer 1

Perhaps some of the "tricky" algebra is what is really tripping you up: you have shown that the base case holds. Now you assume that the statement is true for some fixed $k\geq 1$ (i.e., $\color{blue}{\sum_{i=1}^k i!\cdot i=(k+1)!-1}$), and your goal is to use this assumption (called the inductive hypothesis) to show that $$ \color{green}{\sum_{i=1}^{k+1}i!\cdot i=(k+2)!-1} $$ holds. Start with the left-hand side and work your way to the right-hand side: \begin{align} \color{green}{\sum_{i=1}^{k+1}i!\cdot i} &= \color{blue}{\sum_{i=1}^k i!\cdot i}+(k+1)!\cdot(k+1)\tag{by defn. of $\Sigma$}\\[0.5em] &= \color{blue}{(k+1)!-1}+(k+1)!\cdot(k+1)\tag{by inductive hypothesis}\\[0.5em] &= (k+1)!(1+k+1)-1\tag{factor out $(k+1)!$}\\[0.5em] &= (k+1)!(k+2)-1\tag{simplify}\\[0.5em] &= \color{green}{(k+2)!-1}.\tag{by defn. of factorial} \end{align} Can you see how this proves your claim?

Answer 2

Use induction! By the inductive hypothesis, $$ \sum_{i=1}^{n+1} i! i = \sum_{i=1}^{n} i! i + (n+1)!(n+1) = [(n+1)! - 1] + (n+1)!(n+1) $$

Now just finish off the righthand side.