Induction Proof for a Sum

Question

How would I prove

$$ \sum_{i=1}^n i(i!) = (n+1)! -1 $$

The base case is n = 1, which is trivial, and using the inductive hypothesis, I get to

$$ \sum_{i=1}^{n+1} i(i!) = \sum_{i=1}^n i(i!) + (n+1)(n+1)! $$ which equals $$ (n + 1)! - 1 + (n + 1)(n + 1)! $$ I know it has to end up equaling $$ (k + 2)(k + 1)! - 1\\ $$

but I don't know how to make the jump between the two steps. Any help would be appreciated.

FYI, $(k+2)(k+1)! = (k+2)!$. Also, note $i(i!) = (i+1)(i!) - i! = (i+1)! - i!$, so instead of induction you can use that this is a telescoping series. — John Omielan, Aug 23 '21 at 23:22
Does this answer your question? Proof by induction: $\sum\limits_{i=0}^n i \cdot i! = (n+1)!-1$. Note a comment there links to Summation involving a factorial: $1 + \sum_{j=1}^{n} j!j$ as a possible duplicate, with it having links to other questions about the same problem. — John Omielan, Aug 23 '21 at 23:34

score 4 · Accepted Answer · answered Aug 23 '21 at 23:26

4

I think you can use distributive law.

\begin{align} (n+1)! - 1 + (n + 1)(n + 1)! &= (n+1)! + (n+1)(n+1)! - 1 \\ &= (n+1)! (1 + (n + 1)) - 1 \\ &= (n+1)!(n+2) -1 \end{align}

answered Aug 23 '21 at 23:26

matt

410

Induction Proof for a Sum

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