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    Proof by induction:  \displaystyle\sum_{r=1}^n r(r!) = (n + 1)! - 1

    I'm stuck on the inductive step, the factorials are really messing me up. Could anyone link me to a video which will help me understand how to do these summation proofs with factorials or perhaps explain how I go about solving this?

     (k + 1)! - 1 + (k + 1)[(k + 1)!]

    The form I need somehow get this in is:  (k + 2)! - 1
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    (Original post by FamilyFirst)
    Proof by induction:  \displaystyle\sum_{r=1}^n r(r!) = (n + 1)! - 1

    I'm stuck on the inductive step, the factorials are really messing me up. Could anyone link me to a video which will help me understand how to do these summation proofs with factorials or perhaps explain how I go about solving this?

     (k + 1)! - 1 + (k + 1)[(k + 1)!]

    The form I need somehow get this in is:  (k + 2)! - 1
    Pull out a common factor of (k+1)!:

    (k+1)! - 1 + (k+1)(k+1)! = (k+1)!(1 + k + 1) - 1 = (k+2)(k+1)! - 1

    But you know that k! = k * (k-1) * (k-2) * ... * 1 = k(k-1)! = k(k-1)(k-2)! stc...

    So it makes sense that (k+2)! = (k+2)(k+1)(k)(k-1)...(1) = (k+2)(k+1)!

    So your simplification of (k+2)(k+1)! - 1 becomes...?

    Oh, andsorry for no LaTeX - I'm half aslee and on my phone.
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    (Original post by Zacken)
    ...
    What becomes of (k + 1)! when you multiply it with 1 & k
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    (Original post by FamilyFirst)
    What becomes of (k + 1)! when you multiply it with 1 & k
    1 + k + 1 = k + 2
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    (Original post by Zacken)
    1 + k + 1 = k + 2
    I still don't understand why you did that though :?
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    (Original post by FamilyFirst)
     (k + 1)! - 1 + (k + 1)[(k + 1)!]

    The form I need somehow get this in is:  (k + 2)! - 1
    Okay, I'm on my computer so I can explain this a bit more in depth now. We have: (k+1)! + (k+1)(k+1)! - 1 now we can pull out a common factor of (k+1)!, if you cannot see this, let us call u = (k+1)! so that we have:

    u + (k+1)u -1 then pulling out the common factor of u from the first two terms, we have: u(1 + k + 1) - 1.

    Replacing u by (k+1)! and remembering that 1+k+1 = k+2 we have: (k+2)(k+1)! - 1.

    But (k+2)(k+1)! = (k+2)(k+1)(k)(k-1)\cdots (1) = (k+2)! so we can say that:

    \displaystyle 

\begin{equation*}(k+1)! + (k+1)(k+1)! - 1 = (k+2)(k+1)! - 1 = (k+2)! - 1\end{equation*}
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    (Original post by Zacken)
    But (k+2)(k+1)! = (k+2)(k+1)(k)(k-1)\cdots (1) = (k+2)!
    Yeah that was very smart. I don't think a dopey dude like me would realise that in the instant. Do you think something like this would come up in Edexcel?

    I understood everything you did, but do you know where I can get some extra practice on this, as I don't feel very confident with it?
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    (Original post by FamilyFirst)
    Yeah that was very smart. I don't think a dopey dude like me would realise that in the instant. Do you think something like this would come up in Edexcel?

    I understood everything you did, but do you know where I can get some extra practice on this, as I don't feel very confident with it?
    Aye, this is something that comes with practice - once you see how it's done, you should keep an eye out on it all the time. I'm farily certain it can, has and could/would come up on Edexcel.

    The last two pages on this look good.
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    (Original post by Zacken)
    Aye, this is something that comes with practice - once you see how it's done, you should keep an eye out on it all the time. I'm farily certain it can, has and could/would come up on Edexcel.

    The last two pages on this look good.
    Great,

    thanks!
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    (Original post by FamilyFirst)
    Great,

    thanks!
    Quote me if you need any help on those problems.
 
 
 
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