Prove that $$2\times 2! + 3\times 3!+4\times 4! +....+n\times n!=(n+1)!-2$$ I know that it can be proved by mathematical induction, but I want to prove it without using the mathematical induction. I tied the equation $$C^n_0 + C^n_1 + C^n_2 +...+C^n_n=2^n$$ But I did not get any thing useful.
Asked
Active
Viewed 1,021 times
8
-
5Hint: Telescoping sum. – user202729 Sep 10 '18 at 04:48
-
Are you sure that you have the supposed identity correct? On the left side you have several positive terms being added to $n\times n!$, which is strictly greater than $n!$ which is strictly greater than $(n-1)!$ which is greater still than $(n-1)!-2$... – JMoravitz Sep 10 '18 at 04:49
-
1RHS should be $(n\color{red}{+}1)!-2$ instead. – user202729 Sep 10 '18 at 04:50
-
I am sorry it is as you said that is a misstyping @JMoravitz – Hussien Mohamed Sep 10 '18 at 04:52
2 Answers
17
Hint: $$\sum_{k=2}^n k\cdot k!=\sum_{k=2}^n (k+1-1)\cdot k!=\sum_{k=2}^n (k+1)!-\sum_{k=2}^n k!$$
farruhota
- 32,168
15
$2\times2!=3!-2!$
$3\times3!=4!-3!$
...
$n\times n!=(n+1)!-n!$
Add them all up then $3!,4!,...n!$ cancel out.
Therefore LHS$=(n+1)!-2$
abc...
- 5,010
-
-
-
does it work for any natural number or for $ n \geq 2$ only @abc – Hussien Mohamed Sep 11 '18 at 01:15
-
@HussienMohamed the factorial function is only defined for non-negative integers. – abc... Sep 11 '18 at 01:51
-
i know that , my questions is , does this sum works for any non-negative integer or for any integer $\geq 2$ Only ? – Hussien Mohamed Sep 13 '18 at 00:59
-
-
-
-