# Trying to simplify an expression for an induction proof.

I got it down to $(k+2)!-1 + (k+1)((k+1)!)$ I am trying to get it to $(k+2)!-1$ but I guess I do not understand factorials enough to simplify this. I am also assuming I am doing the induction correctly so far.

prove: that $\sum_{i=1}^n (i)(i!) = (n+1)!-1$ for all positive integers n greater than or equal to 1.

Base Case:

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

LHS = RHS base case holds

Inductive Hypothesis: n = k for arbitrary integer.

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

IS: Show $\sum_{i=1}^{k+1}(i)(i!) = (k+2)!-1$

Proof:

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

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

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

EDIT: added work so far EDIT: cleaned parenthesis up

• Hmm is false... + or - (k+1)...? Apr 1 '16 at 23:03
• I actually made a type I mean i am trying to get it to $(k+2)!-1$ but I am not sure what you're asking.
– Jude
Apr 1 '16 at 23:04
• Can you show us the steps how you got to that formula, and what the problem is? It is unclear from the limited information given where an error was made. Apr 1 '16 at 23:04
• well unless (k+1)((k+1)!)=0 (it doesn't), an error must have been made earlier Apr 1 '16 at 23:06
• @Jude If you put a negative sign before $(k+1)(k+1)!$, your expression would be correct. That is, $(k+2)! - 1 - (k+1)(k+1)! = (k+1)! - 1$ is correct. Apr 1 '16 at 23:07

Your induction hypothesis is $$\color{blue}{\sum_{i = 1}^{k} i \cdot i! = (k + 1)! - 1}$$ You set up the induction step incorrectly. Substituting $k + 1$ for $i$ in the expression $i \cdot i!$ yields $(k + 1)(k + 1)!$. Hence, \begin{align*} \sum_{i = 1}^{k + 1} i \cdot i! & = \color{blue}{\sum_{i = 1}^{k} i \cdot i!} + (k + 1)(k + 1)! && \text{by definition}\\ & = \color{blue}{(k + 1)! - 1} + (k + 1)(k + 1)! && \text{by the induction hypothesis}\\ & = (1 + k + 1)(k + 1)! - 1 && \text{factor}\\ & = (k + 2)(k + 1)! - 1\\ & = (k + 2)! - 1 \end{align*}