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Minor question, but cannot seem to solve

The first part of this question is:

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

So I started with:

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

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

Then surely

r+1r!(r+1)!≡r(r+1)!\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.
Amateur mistakes.

Edit: Just Kidding people.
Original post by Femto
The first part of this question is:

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

So I started with:

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

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

Then surely

r+1r!(r+1)!≡r(r+1)!\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 1r!−1(r+1)! \dfrac{1}{r!} - \dfrac{1}{(r+1)!} actually means, you'll find that

1r!−1(r+1)!=1r.(r−1)...(2).(1)−1(r+1).(r).(r−1)...(2).(1) \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?
Reply 3
Avatar for BJack
BJack
19
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!≡r![(r+1)−1](r+1)! - r! \equiv r![(r+1)-1]. Do you see where to go from here?
Reply 4
Avatar for Femto
Femto
OP
8
Original post by Straight up G
Amateur mistakes.

Edit: Just Kidding people.


You're probably right to be honest :s-smilie:
Reply 5
Avatar for Femto
Femto
OP
8
Original post by Clarity Incognito
Overcomplicating it a touch.

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

1r!−1(r+1)!=1r.(r−1)...(2).(1)−1(r+1).(r).(r−1)...(2).(1) \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?
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.
Reply 7
Avatar for Femto
Femto
OP
8
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?

(r+1)−1r!\frac{(r + 1) - 1}{r!}
Reply 8
Avatar for BJack
BJack
19
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?

(r+1)−1r!\frac{(r + 1) - 1}{r!}


Did you see my post?
(edited 13 years ago)
Reply 9
Avatar for Femto
Femto
OP
8
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 :biggrin: I am doing this wrong right?
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!
Reply 11
Avatar for BJack
BJack
19
Original post by Femto
Yes I am aware of it, thanks very much for your reply :biggrin: I am doing this wrong right?


You should probably write it out as

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

rather than

⋯≡r(r+1)!\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.
Reply 12
Avatar for Femto
Femto
OP
8
Original post by BJack
You should probably write it out as

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

rather than

⋯≡r(r+1)!\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 :s-smilie:
Reply 13
Avatar for BJack
BJack
19
Original post by Femto
Yeah I see what you mean, but I'm struggling to go from where I left off in the OP :s-smilie:


(r+1)!−r!r!(r+1)!≡r!(r+1)−r!r!(r+1)!\displaystyle \frac{(r+1)! - r!}{r!(r+1)!} \equiv \frac{r!(r+1) - r!}{r!(r+1)!}
Reply 14
Avatar for awais590
awais590
12
Original post by Straight up G
Amateur mistakes.

Edit: Just Kidding people.


Noob.
Original post by awais590
Noob.


What you talkin bout waisy you dont talk to Don like that
Reply 16
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awais590
12
Original post by Straight up G
What you talkin bout waisy you dont talk to Don like that


:tongue:
Reply 17
Avatar for Femto
Femto
OP
8
Original post by BJack
(r+1)!−r!r!(r+1)!≡r!(r+1)−r!r!(r+1)!\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!

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