Turn on thread page Beta
    • Thread Starter
    Offline

    8
    ReputationRep:
    The first part of this question is:

    Show that \frac{1}{r!} - \frac{1}{(r + 1)!} \equiv \frac{r}{(r + 1)!}

    So I started with:

    \frac{(r + 1)! - r!}{r!(r + 1)!} \equiv \frac{r}{(r + 1)!}

    \frac{(r + 1)(r)(r - 1)\times... - (r)(r - 1)\times... }{r!(r + 1)!} \equiv \frac{r}{(r + 1)!}

    Then surely

    \frac{r + 1}{r!(r + 1)!} \equiv \frac{r}{(r + 1)!}

    I'd go further but, this seems incorrect - could anybody guide me through this? I hate dealing with the factorial.
    Offline

    0
    ReputationRep:
    Amateur mistakes.

    Edit: Just Kidding people.
    Offline

    10
    ReputationRep:
    (Original post by Femto)
    The first part of this question is:

    Show that \frac{1}{r!} - \frac{1}{(r + 1)!} \equiv \frac{r}{(r + 1)!}

    So I started with:

    \frac{(r + 1)! - r!}{r!(r + 1)!} \equiv \frac{r}{(r + 1)!}

    \frac{(r + 1)(r)(r - 1)\times... - (r)(r - 1)\times... }{r!(r + 1)!} \equiv \frac{r}{(r + 1)!}

    Then surely

    \frac{r + 1}{r!(r + 1)!} \equiv \frac{r}{(r + 1)!}

    I'd go further but, this seems incorrect - could anybody guide me through this? I hate dealing with the factorial.
    Overcomplicating it a touch.

    If you think about what  \dfrac{1}{r!} - \dfrac{1}{(r+1)!} actually means, you'll find that

     \dfrac{1}{r!} - \dfrac{1}{(r+1)!} = \dfrac{1}{r.(r-1)...(2).(1)} - \dfrac{1}{(r+1).(r).(r-1)...(2).(1)}

    How do you get a common denominator now?
    Offline

    19
    ReputationRep:
    (Original post by Femto)
    I'd go further but, this seems incorrect - could anybody guide me through this? I hate dealing with the factorial.
    (r+1)! - r! \equiv r![(r+1)-1]. Do you see where to go from here?
    • Thread Starter
    Offline

    8
    ReputationRep:
    (Original post by Straight up G)
    Amateur mistakes.

    Edit: Just Kidding people.
    You're probably right to be honest
    • Thread Starter
    Offline

    8
    ReputationRep:
    (Original post by Clarity Incognito)
    Overcomplicating it a touch.

    If you think about what  \dfrac{1}{r!} - \dfrac{1}{(r+1)!} actually means, you'll find that

     \dfrac{1}{r!} - \dfrac{1}{(r+1)!} = \dfrac{1}{r.(r-1)...(2).(1)} - \dfrac{1}{(r+1).(r).(r-1)...(2).(1)}

    How do you get a common denominator now?
    Hmm, I'm sorry I cannot see what the common denominator would be unless it's r!? Would you mind explaining to me how you got the (2).(1) in the denominator?
    Offline

    10
    ReputationRep:
    (Original post by Femto)
    Hmm, I'm sorry I cannot see what the common denominator would be unless it's r!? Would you mind explaining to me how you got the (2).(1) in the denominator?
    From the definition of a factorial, 1!=1, 2!=2x1, 3!=3x2x1, ...., n!=nx(n-1)x(n-2)x...x3x2x1.

    Yeah, they both have r! in the denominator and you want to muultiply one of the fractions by something else so that they both have the same denominator, then you simplify your expression.
    • Thread Starter
    Offline

    8
    ReputationRep:
    (Original post by Clarity Incognito)
    From the definition of a factorial, 1!=1, 2!=2x1, 3!=3x2x1, ...., n!=nx(n-1)x(n-2)x...x3x2x1.

    Yeah, they both have r! in the denominator and you want to muultiply one of the fractions by something else so that they both have the same denominator, then you simplify your expression.
    Would it become?

    \frac{(r + 1) - 1}{r!}
    Offline

    19
    ReputationRep:
    (Original post by Clarity Incognito)
    From the definition of a factorial, 1!=1, 2!=2x1, 3!=3x2x1, ...., n!=nx(n-1)x(n-2)x...x3x2x1.

    Yeah, they both have r! in the denominator and you want to muultiply one of the fractions by something else so that they both have the same denominator, then you simplify your expression.
    This assumes that you're dealing with integer factorials! :p:

    (Original post by Femto)
    Would it become?

    \frac{(r + 1) - 1}{r!}
    Did you see my post?
    • Thread Starter
    Offline

    8
    ReputationRep:
    (Original post by BJack)
    This assumes that you're dealing with integer factorials! :p:



    Did you see my post?
    Yes I am aware of it, thanks very much for your reply I am doing this wrong right?
    Offline

    10
    ReputationRep:
    (Original post by BJack)
    This assumes that you're dealing with integer factorials! :p:
    Haha, ok, my method was purely for positive integers but it can be easily adapted nonetheless to produce the same result! Good shout nonetheless, it just looked like an A level question to me!
    Offline

    19
    ReputationRep:
    (Original post by Femto)
    Yes I am aware of it, thanks very much for your reply I am doing this wrong right?
    You should probably write it out as

    \displaystyle \frac{1}{r!} - \frac{1}{(r+1)!} \equiv \frac{(r+1)! -r!}{r!(r+1)!} \equiv \dots

    rather than

    \dots \equiv \displaystyle \frac{r}{(r+1)!}

    for each stage. Apart from that, your approach is fine, though expanding the factorials is unnecessary when you can just factorize the resulting numerator almost immediately.
    • Thread Starter
    Offline

    8
    ReputationRep:
    (Original post by BJack)
    You should probably write it out as

    \displaystyle \frac{1}{r!} - \frac{1}{(r+1)!} \equiv \frac{(r+1)! -r!}{r!(r+1)!} \equiv \dots

    rather than

    \dots \equiv \displaystyle \frac{r}{(r+1)!}

    for each stage. Apart from that, your approach is fine, though expanding the factorials is unnecessary when you can just factorize the resulting numerator almost immediately.
    Yeah I see what you mean, but I'm struggling to go from where I left off in the OP
    Offline

    19
    ReputationRep:
    (Original post by Femto)
    Yeah I see what you mean, but I'm struggling to go from where I left off in the OP
    \displaystyle \frac{(r+1)! - r!}{r!(r+1)!} \equiv \frac{r!(r+1) - r!}{r!(r+1)!}
    Offline

    11
    ReputationRep:
    (Original post by Straight up G)
    Amateur mistakes.

    Edit: Just Kidding people.
    Noob.
    Offline

    0
    ReputationRep:
    (Original post by awais590)
    Noob.
    What you talkin bout waisy you dont talk to Don like that
    Offline

    11
    ReputationRep:
    (Original post by Straight up G)
    What you talkin bout waisy you dont talk to Don like that
    • Thread Starter
    Offline

    8
    ReputationRep:
    (Original post by BJack)
    \displaystyle \frac{(r+1)! - r!}{r!(r+1)!} \equiv \frac{r!(r+1) - r!}{r!(r+1)!}
    Aah yes of course!! Thank you very much!
 
 
 
Poll
How are you feeling in the run-up to Results Day 2018?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.