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\begin{align*} \sum_{i = 1}^{k + 1} i(i!) & = \sum_{i = 1}^{k} i(i!) + (k + 1)(k + 1)!\\ & = (k + 1)! - 1 + (k + 1)(k + 1)! & \text{by the induction hypothesis}\\ & = (1 + k + 1)(k + 1)! - 1\\ & = (k + 2)(k + 1)! - 1\\ & = (k + 2)! - 1 \end{align*}

I have a question from this post solving the problem Prove by induction that $\sum_{i=1}^n i!\times i=(n+1)!-1$ for all $n\in \mathbb{N}$

How does the person go from $ = (k + 2)(k + 1)! - 1\\ = (k + 2)! - 1$

at the very end? I don't understand how the permutation of $(k+1)!$ and (k+2) are able to combine into $(k+2)!$

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    $\begingroup$ Consider an example, say with $k=3$: $$(k+2)(k+1)!=(3+2)(3+1)!=5\cdot4!=5(4\cdot3\cdot2\cdot1)=5\cdot4\cdot3\cdot2\cdot1=5!=(3+2)!=(k+2)!$$ $\endgroup$ Feb 6, 2019 at 15:16
  • $\begingroup$ ah that makes sense, thank you $\endgroup$
    – Evan Kim
    Feb 6, 2019 at 22:51

2 Answers 2

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It is because, by definition, $n! = n(n-1)!$ and $0! = 1$. Just take $n=k+2$.

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Why not showing it directly?

\begin{eqnarray*} \sum_{i=1}^{n} i\cdot i! & = & \sum_{i=1}^{n} (i+1-1)\cdot i! \\ & = & \sum_{i=1}^{n} \left((i+1)!- i!\right) \\ & \stackrel{telescoping}{=} & (n+1)! - 1\\ \end{eqnarray*}

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    $\begingroup$ The OP might wonder how you got from $(i+1)i!$ to $(i+1)!$.... $\endgroup$ Feb 6, 2019 at 15:13

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