I'm trying to prove $$ \sum_{i=1}^n i.i! = (n+1)!-1 $$ with mathematical induction. The first step I did after prove it for 1 was: $$ \sum_{i=1}^{n+1} i.i! = (n+2)!-1= (n+2).(n+1)!-1 $$ but I can't do anything more.
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Sahiba Arora
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Amir reza Riahi
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The answer by mfl at the duplicate is by induction. – Dietrich Burde Jun 04 '20 at 11:58
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@DietrichBurde how can I find my similar question here when I don't have any keyword? just like this question? – Amir reza Riahi Jun 04 '20 at 12:01
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You were almost there btw, $$([n+1]+1)!-1$$ was what you had to split it into to prove it holds for $n+1$ – sai-kartik Jun 04 '20 at 12:02
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@DietrichBurde, you can also try using Approach0 to search for similar questions – sai-kartik Jun 04 '20 at 12:04
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The equality is trivial for $n=1.$ Assume it is true for $n,$ then $$\begin{align*}\sum_{i=1}^{n+1} i\cdot i!&=\sum_{i=1}^n i.i!+(n+1).(n+1)!\\&=(n+1)!-1+(n+1)(n+1)! \qquad \text{(induction hypothesis)}\\ &=(n+2)(n+1)!-1\\&=(n+2)!-1.\end{align*}$$
Sahiba Arora
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