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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

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  • $\begingroup$ Hmm is false... + or - (k+1)...? $\endgroup$ – Martín Vacas Vignolo Apr 1 '16 at 23:03
  • $\begingroup$ 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. $\endgroup$ – Jude Apr 1 '16 at 23:04
  • $\begingroup$ 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. $\endgroup$ – Chill2Macht Apr 1 '16 at 23:04
  • $\begingroup$ well unless (k+1)((k+1)!)=0 (it doesn't), an error must have been made earlier $\endgroup$ – Chill2Macht Apr 1 '16 at 23:06
  • $\begingroup$ @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. $\endgroup$ – Deepak Apr 1 '16 at 23:07
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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*}

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  • $\begingroup$ I am actually a bit confused on what/how you factored out the factorial $\endgroup$ – Jude Apr 1 '16 at 23:45
  • $\begingroup$ Observe that \begin{align*}(k + 1)! - 1 + (k + 1)(k + 1)! & = \color{green}{1} \cdot (k + 1)! + \color{green}{(k + 1)}(k + 1)! - 1\\ & = \color{green}{(1 + k + 1)}(k + 1)! - 1\end{align*} $\endgroup$ – N. F. Taussig Apr 2 '16 at 0:11

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