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Use induction to prove that $\displaystyle\sum_{r=1}^n r\cdot r! =(n+1)! -1$

I first showed that the formula holds true for $n=1$. Then I put n as $k$ and got an expression for the sum in terms of $k$. I then found the sum till the $(k+1)$th term by adding the $(k+1)$th term to both sides of the equation and compared it to the sum expression I get by plugging $k+1$ into the sum expression given in the question - they don't match.

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  • $\begingroup$ TeX tip: use \cdot for centered dot, as in $2\cdot3=6$. $\endgroup$ – user147263 Sep 2 '14 at 19:43
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HINT

$$\begin{align} \\ (k+1)! -1 + (k+1)(k+1)! &= (k+1)!(1 + k+1) - 1 \\&= (k+1)!(k+2)-1\\&= ? \end{align}$$

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  • $\begingroup$ That's exactly what i'm doing! It simplifies to (k+2)!(k+1)! -1 $\endgroup$ – user140161 Sep 2 '14 at 19:28
  • $\begingroup$ I see... so the proof is complete, right ? $\endgroup$ – ganeshie8 Sep 2 '14 at 19:29
  • $\begingroup$ Oh wait a second, how does that simplify to (k+2)!(k+1)!-1 ? $$n!(n+1) = (n+1)!$$ $\endgroup$ – ganeshie8 Sep 2 '14 at 19:31
  • $\begingroup$ shouldnt the sum be (k+2)! - 1? I am sorry if my questions seem silly, I just started induction. $\endgroup$ – user140161 Sep 2 '14 at 19:31
  • $\begingroup$ yes, and thats exactly what you need to conclude the statement holds for $n = k+1$ because $$(k+2)! - 1 = (\color{Red}{k+1} + 1)! - 1$$ $\endgroup$ – ganeshie8 Sep 2 '14 at 19:33
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Also a neat way to solve it without induction $$\sum_{r=1}^nr\cdot r!=\sum_{r=1}^{n}(r+1-1)\cdot r!=\sum_{r=1}^n (r+1)!-r!=(n+1)!-1$$ Since all terms cancel out except $(n+1)!$ and $-1$

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  • $\begingroup$ I agree with you that your way is neat :). +1 $\endgroup$ – Chinny84 Sep 2 '14 at 20:12

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