# 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)...? – Martín Vacas Vignolo 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. – Chill2Macht Apr 1 '16 at 23:04
• well unless (k+1)((k+1)!)=0 (it doesn't), an error must have been made earlier – Chill2Macht 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. – Deepak 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*}