Question

Use induction to prove that for all integers n 2 1 we have 1.1! +2.2! +3.3! + ... +nin! = (n + 1)! - 1.

Answer 1

Let $P(n)$ be the statement that

$1\cdot1!+2\cdot2!+\cdots+n\cdot n!=(n+1)!-1$

For $n=1$, we have $P(1)$ true, since

$1\cdot1!=1$

and

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

Assume $P(k)$ is true. Then

$1\cdot1!+2\cdot2!+\cdots+k\cdot k!=(k+1)!-1$

Then for $n=k+1$ we have

$1\cdot1!+2\cdot2!+\cdots+k\cdot k!+(k+1)\cdot(k+1)!=(k+1)!-1+(k+1)\cdot(k+1)!$

because $P(k)$ is assumed to be true, then

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

and so $P(n)$ is true for all $n$. QED

Alternatively, no induction is needed, since

$(n+1)!=(n+1)\cdot n!=n\cdot n!+n!$

so

$1\cdot1!+2\cdot2!+\cdots+n\cdot n!=(2!-1!)+(3!-2!)+\cdots+((n+1)!-n!)=(n+1)!-1!$

$\implies1\cdot1!+2\cdot2!+\cdots+n\cdot n!=(n+1)!-1$