0
$\begingroup$

I'm trying to prove $$ \sum_{i=1}^n i.i! = (n+1)!-1 $$ with mathematical induction. The first step I did after prove it for 1 was: $$ \sum_{i=1}^{n+1} i.i! = (n+2)!-1= (n+2).(n+1)!-1 $$ but I can't do anything more.

$\endgroup$
4
  • $\begingroup$ The answer by mfl at the duplicate is by induction. $\endgroup$ Jun 4, 2020 at 11:58
  • $\begingroup$ @DietrichBurde how can I find my similar question here when I don't have any keyword? just like this question? $\endgroup$ Jun 4, 2020 at 12:01
  • $\begingroup$ You were almost there btw, $$([n+1]+1)!-1$$ was what you had to split it into to prove it holds for $n+1$ $\endgroup$
    – sai-kartik
    Jun 4, 2020 at 12:02
  • $\begingroup$ @DietrichBurde, you can also try using Approach0 to search for similar questions $\endgroup$
    – sai-kartik
    Jun 4, 2020 at 12:04

1 Answer 1

1
$\begingroup$

The equality is trivial for $n=1.$ Assume it is true for $n,$ then $$\begin{align*}\sum_{i=1}^{n+1} i\cdot i!&=\sum_{i=1}^n i.i!+(n+1).(n+1)!\\&=(n+1)!-1+(n+1)(n+1)! \qquad \text{(induction hypothesis)}\\ &=(n+2)(n+1)!-1\\&=(n+2)!-1.\end{align*}$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .