Question

For all $n\in N,1\times 1!+2\times 2!+3\times 3!+\cdots +n\times n!$ is equal to

• A. $(n+1)!-2$
• B. $(n+1)!$
• C. $(n+1)!-1$
• D. $(n+1)!-3$

Answer 1

Answer: (C)

Let the statement $P(n)$ be defined as
$P(n):1\times 1!+2\times 2!+3\times 3!\dots +n\times n!=(n+1)!-1$ for all natural numbers $n$.
Note that $P(1)$ is true, since
$P(1):1\times 1!=1=2-1=2!-1$
Assume that $P(n)$ is true for some natural number $k$,
i.e., $P(k):1\times 1!+2\times 2!+3\times 3!+\dots +k\times k!=(k+1)!-\mathrm{1...}$(i)
To prove $P(k+1)$ is true, we have
$P(k+1):1\times 1!+2\times 2!+3\times 3!+\cdots +k\times k!+(k+1)\times (k+1)!$
$=(k+1)!-1+(k+1)!\times (k+1)\text{ [by Eq. }(i)]$
$=(k+1+1)(k+1)!-1$
$=(k+2)(k+1)!-1$
$=(k+2)!-1$
Thus, $P(k+1)$ is true, whenever $P(k)$ is true. Therefore, by the principle of mathematical induction, $P(n)$ is true for all natural numbers $n$.