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    whats the lowest common denominator of this factorial and why? thanks

    this isn't equivalent to: 1/x - 1/(x+1)

    so how would I go about it?
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    Write the factorials out explicitly and it should become obvious.
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    (Original post by IrrationalRoot)
    Write the factorials out explicitly and it should become obvious.
    I did try that at first but now I see it
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    (Original post by IrrationalRoot)
    Write the factorials out explicitly and it should become obvious.
    so, I see how the lowest common denominator is (r+1)!

    however, I dont understand the first fraction. how does: r!(r+1) = (r+1)!?
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    (Original post by Maths&physics)
    so, I see how the lowest common denominator is (r+1)!

    however, I dont understand the first fraction. how does: r!(r+1) = (r+1)!?
     \displaystyle n! = n \cdot \underbrace{(n-1) \cdot (n-2) \cdot ... \cdot 2 \cdot 1}_{= (n-1)!}
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    (Original post by RDKGames)
     \displaystyle n! = n \cdot \underbrace{(n-1) \cdot (n-2) \cdot ... \cdot 2 \cdot 1}_{= (n-1)!}
    sorry but I dont get what you wrote.

    r!(r+1) = (r+1)!

    1(1+1) x 2(2+1) x 3(3+1) = (1+1) x (2+1) x (3+1)

    2 x 6 x 12 = 2 x 3 x 4 ????
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    (Ok the app is being useless so I cant quote you properly)

    Anyway, what do you think ‘!’ means?
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    (Original post by Maths&physics)
    sorry but I dont get what you wrote.

    r!(r+1) = (r+1)!

    1(1+1) + 2(2+1) + 3(3+1) = (1+1) + (2+1) + (3+1)

    2 + 6 + 12 = 2 + 3 + 4 ????
    You're overcomplicating this, it's just:

    r!(r+1)=(1\times 2\times 3\times \ldots \times r) \times (r+1)= 1\times 2\times 3\times \ldots \times r \times (r+1) = (r+1)!.

    As you can see it follows immediately from the definition of factorial (actually it IS the definition of factorial, technically speaking).
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    (Original post by RDKGames)
    (Ok the app is being useless so I cant quote you properly)

    Anyway, what do you think ‘!’ means?
    I thought ! = 1x2x3x4, etc?
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    (Original post by Maths&physics)
    I thought ! = 1x2x3x4, etc?
    Yes.... n!=n(n-1)(n-2)...(2)(1) as I said in my initial post.

    You didn’t apply this definition in your flawed example
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    (Original post by IrrationalRoot)
    You're overcomplicating this, it's just:

    r!(r+1)=(1\times 2\times 3\times \ldots \times r) \times (r+1)= 1\times 2\times 3\times \ldots \times r \times (r+1) = (r+1)!.

    As you can see it follows immediately from the definition of factorial (actually it IS the definition of factorial, technically speaking).

    so its:

    (1x2x3x4....)(r+1) = (1+1)(1+2)(1+3)(1+4)....

    (1x2x3x4....)(1+1)(2+1)(3+1)(4+1 ) = (1+1)(1+2)(1+3)(1+4)....

    I dont see how theyre equivalent?
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    (Original post by IrrationalRoot)
    You're overcomplicating this, it's just:

    r!(r+1)=(1\times 2\times 3\times \ldots \times r) \times (r+1)= 1\times 2\times 3\times \ldots \times r \times (r+1) = (r+1)!.

    As you can see it follows immediately from the definition of factorial (actually it IS the definition of factorial, technically speaking).
    ah, do you automatically multiply whats inside the bracket by 1x2x3.... when there is an ! besides it?

    so r! = (1x2x3....)(1x2x3...)?
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    (Original post by RDKGames)
    Yes.... n!=n(n-1)(n-2)...(2)(1) as I said in my initial post.

    You didn’t apply this definition in your flawed example
    I didn't: sorry
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    (Original post by Maths&physics)
    so its:

    (1x2x3x4....)(r+1) = (1+1)(1+2)(1+3)(1+4)....

    (1x2x3x4....)(1+1)(2+1)(3+1)(4+1 ) = (1+1)(1+2)(1+3)(1+4)....

    I dont see how theyre equivalent?
    (Original post by Maths&physics)
    ah, do you automatically multiply whats inside the bracket by 1x2x3.... when there is an ! besides it?

    so r! = !

    its the same thing?
    ??????????

    No! We are clearly saying that whenever you see something with a ! right after it, that means you multiply every integer from 1 up to it together..
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    (Original post by Maths&physics)
    ah, do you automatically multiply whats inside the bracket by 1x2x3.... when there is an ! besides it?

    so r! = !

    its the same thing?
    1!=1

    2!=1\times 2

    3!=1 \times 2 \times 3

    4!=1 \times 2 \times 3 \times 4

    r!=1 \times 2 \times 3 \times 4 \times \ldots \times (r-1) \times r

    You really can't go wrong with this.
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    (Original post by RDKGames)
    ??????????

    No! We are clearly saying that whenever you see something with a ! right after it, that means you multiply every integer from 1 up to it together..
    so r! = (1x2x3..)(1x2x3...)?
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    (Original post by Maths&physics)
    I didn't: sorry
    You seem confused in these posts. Taking it back to basics do you know what 5! means? And do you understand why these are true?

    5! = 5 \times 4!

    5! = 5\times 4 \times 3!

    I think it's best that you understand this before introducing variables.
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    (Original post by Maths&physics)
    so r! = (1x2x3..)(1x2x3...)?
    Just open up your textbook and look at how it’s defined and used.
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    OK, I'VE FINALLY GOT IT! thanks
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    (Original post by IrrationalRoot)
    You're overcomplicating this, it's just:

    r!(r+1)=(1\times 2\times 3\times \ldots \times r) \times (r+1)= 1\times 2\times 3\times \ldots \times r \times (r+1) = (r+1)!.

    As you can see it follows immediately from the definition of factorial (actually it IS the definition of factorial, technically speaking).
    ive got it.

    r = 3

    3!(3+1) = (3+1)!

    I just realised you told me that about an hour ago!


    thanks
 
 
 
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