Question

Prove by the principle of mathematical induction that $1\times 1!+2\times 2!+3\times 3!+...+n\times n!=(n+1)!-1$ for all natural numbers n.

Answer 1

Let P(n) be the given statement, i.e.

$P\left(n\right):1\times 1!+2\times 2!+3\times 3!+...+n\times n!=(n+1)!-1$

For n = 1,

$P\left(1\right):1\times 1!=(1+1)!-1\Rightarrow 1=2!-1\Rightarrow 1=1$

Hence, P(1) is true.

Assume that, P(n) is true for some natural number k, i.e.

$P\left(k\right):1\times 1!+2\times 2!+3\times 3!+...+n\times n!=(n+1)!-1$

We will now show that P(k + 1) is true, whenever

P(k) is true. For this we have to show that

$1\times 1!+2\times 2!+3\times 3!+...+k\times k!(k+1)\times (k+1)!$

= $(k+1+1)!-1$

Now, adding $(k+1)\times (k+1)!$ both sides of Eq. (i), we get

$1\times 1!+2\times 2!+3\times 3!+...+k\times k!+(k+1)\times (k+1)!$

= $(k+1)!-1+(k+1)\times (k+1)!$

= $(k+1)!+(k+1)\times (k+1)!-1$

= $(k+1)!(1+k+1)-1=(k+2)(k+1)!-1$

= $(k+2)!-1=(k+1+1)!-1$

$\therefore P(k+1)$ is true.

Hence, by principle of mathematical induction P(n) is true of all $n\in N.$