Chapter 5, Problem 1RE

Use mathematical induction to show that ${\displaystyle \underset{i=1}{\overset{n}{\sum }}i\left(i!\right)}=\left(n+1\right)!-1$.

To determine

To prove: That ${\displaystyle \sum _{i=1}^{n}i\left(i!\right)}=\left(n+1\right)!-1$ using mathematical induction.

Explanation of Solution

Proof:

For $n=1,$

$\begin{array}{l}1\left(1!\right)=1\\ \left(1+1\right)!-1=2!-1=1\end{array}$

Thus, the result is true for $n=1$.

Now assume that the result is true for $n=k$.

That is

${\displaystyle \sum _{i=1}^{k}i\left(i!\right)}=\left(k+1\right)!-1$

We will now prove that the result is true for $n=k+1$.

That is ${\displaystyle \sum _{i=1}^{k+1}i\left(i!\right)}=\left(k+1+1\right)!-1=\left(k+2\right)!-1$.

Now,

$\begin{array}{c}{\displaystyle \sum _{i=1}^{k+1}i\left(i!\right)}={\displaystyle \sum _{i=1}^{k}i\left(i!\right)}+(k+1)\left(k+1\right)!\\ =\left(k+1\right)!-1+(k+1)\left(k+1\right)!\\ =\left(k+1\right)!\left(1+(k+1)\right)-1\\ =\left(k+1\right)!\left(k+2\right)-1\\ =\left(k+2\right)!-1\end{array}$

Thus, the result is true for $n=k+1$.

Hence, by the principle of mathematical induction, the result is true for all $n\ge 1$.