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Expand the summation $$\sum^{n}_{i=0}i\times i! = $$

I am studying for an exam, I have no idea what this question means.

juantheron
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Doctor X
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  • The question has many details. Do you understand the symbols used in the representation of the "summation"? – Math-fun Oct 22 '15 at 07:54
  • Yes, summation (sigma) is the symbol which adds all the following numbers together, but it is the i x i! which i am confused about, is it (n(n+1)) / 2 . (n!(n!+1)/ 2? and does that lead to( n^2 + n + n!^2 + n! )/ 2? – Doctor X Oct 22 '15 at 08:07
  • lets define a function $f(i) = i \times i !$. now let'S evaluate this function at different values to see how does that behave: $f(1)=1 \times 1 ! = 1 \times 1 =1$, $f(2)=2 \times 2 ! = 2 \times (2\times 1) =4$, $f(3)=3 \times 3 ! = 3 \times (3 \times 2 \times 1) =18$ and so on. Does this make the function clear? – Math-fun Oct 22 '15 at 08:12
  • I understand factorial notation, but I do not understand where it ends when expanding the summation, I am just going around in a circle. – Doctor X Oct 22 '15 at 08:16
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    the summation ends at $n$, hence the last term that we should to the sum is $n \times n!$. Does this answer your question? – Math-fun Oct 22 '15 at 08:27
  • Possible duplicate of Prove by Mathematical Induction: $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$ –  Oct 22 '15 at 13:27

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Let $$\displaystyle \bf{S_{n} = \sum^{n}_{i=0}i!\times i = \sum^{n}_{i=0}i!\times [(i+1)-1]} = \sum^{n}_{i=0}\left[i!\times (i+1)-i!\right]$$

So we get $$\displaystyle \bf{S_{n} = \sum^{n}_{i=0}\left[(i+1)!-i!\right]}$$

Now Expanding Summation or Using Telescopic Sum , we get

$$\displaystyle \bf{S_{n} = \sum^{n}_{i=0}\left[(i+1)!-i!\right]} = (n+1)!-1$$

juantheron
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    The user does not know what the summation means. And you are asking him to expand using telescopic sums? – Apurv Oct 22 '15 at 07:42