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Prove the summation: $$\sum_{i=1}^n i! \cdot i = (n+1)! - 1$$ using induction.

base case: $n=1$: \begin{align*} \sum_{i=1}^1 i! \cdot i &= (1+1)! - 1 \\ 1 &= 2 - 1 \\ 1 &= 1 \end{align*}

This is a question from my test review packet, currently have the base case completed and I am a bit lost on where to go from there. Any help/hints are appreciated.

Martin Sleziak
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D.Peterson
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    I know you have to prove this by induction, but this is one of those things that's just so easy to prove without it. Note that $i!\cdot i = (i+1)! - i!$. – Deepak Sep 29 '15 at 00:37
  • A few related potsts: http://math.stackexchange.com/questions/301615/prove-by-mathematical-induction-11-22-cdot-cdot-cdot-nn-n http://math.stackexchange.com/questions/326105/sum-i-1n-i-cdot-i-n1-1-by-induction http://math.stackexchange.com/questions/410290/evaluate-sum-k-1nk-cdot-k http://math.stackexchange.com/questions/917367/using-induction-to-prove-that-sum-r-1n-r-cdot-r-n1-1 http://math.stackexchange.com/questions/948757/prove-11-dotsnn-n1-1-using-induction – Martin Sleziak Oct 04 '16 at 00:34

2 Answers2

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Perhaps some of the "tricky" algebra is what is really tripping you up: you have shown that the base case holds. Now you assume that the statement is true for some fixed $k\geq 1$ (i.e., $\color{blue}{\sum_{i=1}^k i!\cdot i=(k+1)!-1}$), and your goal is to use this assumption (called the inductive hypothesis) to show that $$ \color{green}{\sum_{i=1}^{k+1}i!\cdot i=(k+2)!-1} $$ holds. Start with the left-hand side and work your way to the right-hand side: \begin{align} \color{green}{\sum_{i=1}^{k+1}i!\cdot i} &= \color{blue}{\sum_{i=1}^k i!\cdot i}+(k+1)!\cdot(k+1)\tag{by defn. of $\Sigma$}\\[0.5em] &= \color{blue}{(k+1)!-1}+(k+1)!\cdot(k+1)\tag{by inductive hypothesis}\\[0.5em] &= (k+1)!(1+k+1)-1\tag{factor out $(k+1)!$}\\[0.5em] &= (k+1)!(k+2)-1\tag{simplify}\\[0.5em] &= \color{green}{(k+2)!-1}.\tag{by defn. of factorial} \end{align} Can you see how this proves your claim?

Daniel W. Farlow
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  • I see it now. Thanks a ton! I'm still a bit new to induction so I think the use of the inductive hypothesis is what was tripping me up. Thanks again! – D.Peterson Sep 29 '15 at 18:47
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Use induction! By the inductive hypothesis, $$ \sum_{i=1}^{n+1} i! i = \sum_{i=1}^{n} i! i + (n+1)!(n+1) = [(n+1)! - 1] + (n+1)!(n+1) $$

Now just finish off the righthand side.

Michael Knopf
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