induction with factorials

Question

I need help with this please. I understand step one is to let $n=1$. step two let $ n = k$. Step three prove for $k+1$. But I would like a clear example of each...

Prove

$$\sum_{i=1}^n i(i!)=(n+1)!−1 $$

Answer 1

I wouldn't divide steps two and three; they are the same move. After step one (the base case) you assume that your equality holds for all positive integers $n$ where $1<n \leq k$ for SOME arbitrary $k$. This is called the induction step, and it is at this point that you want to tinker around with your equality evaluated at $(k+1)$. You should be able to prove the induction step on your own. You simply show that the equality $$\sum_{i=1}^n i(i!)=(n+1)!−1 $$ holds for $n=1$. Once you have done this, then consider the quantity $$\sum_{i=1}^{k+1} i(i!) = (k+1)(k+1)!+k(k!)+(k-1)(k-1)!+ \dots+1(1)!$$ If you mess around with the expanded sum, you should be able to prove that $$(k+1)(k+1)!+k(k!)+(k-1)(k-1)!+ \dots+1(1)!= (k+2)!-1$$ and hence $$\sum_{i=1}^{k+1} i(i!) = (k+2)!-1$$ At this point, you will have completed the proof by induction.