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Math Problem 2)

Posted: Fri Oct 23, 2009 5:47 pm
by penfold1992
alright i dno where to start... this is probably much harder then what Crimson posted but still

Show that 1/r! - 1/(r+1)! = r/(r+1)!

and if some1 asks. r is any number... r doesnt equal something specifically.

if you need another clue i know it has something to do with proof by induction

good luck :)

thanks in advance too this is in meh homework for half term
EDIT:

HENCE find an expression, in terms of n, for
1/2! + 2/3! +3/4!+......+n/(n+1)!

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 6:04 pm
by CeLL
I havent been in school in a while whats "!" stand for?

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 6:08 pm
by EvGa
! is factorial. So 4! = 4*3*2*1.

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 6:14 pm
by CeLL
o ****, have fun man.

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 6:29 pm
by only_me
penfold1992 wrote: Show that 1/r! - 1/(r+1)! = r/(r+1)!


multyply 1/r! with r+1 and u ll have r+1/(r+1)!

and r+1/(r+1)! - 1/(r+1)! its r/(r+1)! . simple no ?:O

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 6:50 pm
by Sylar
only_me wrote:
penfold1992 wrote: Show that 1/r! - 1/(r+1)! = r/(r+1)!


multyply 1/r! with r+1 and u ll have r+1/(r+1)!

and r+1/(r+1)! - 1/(r+1)! its r/(r+1)! . simple no ?:O


if you multiply 1/r! with r+1 you'll get r+1/r! , no ?

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 6:51 pm
by penfold1992
only_me wrote:
penfold1992 wrote: Show that 1/r! - 1/(r+1)! = r/(r+1)!


multyply 1/r! with r+1 and u ll have r+1/(r+1)!

and r+1/(r+1)! - 1/(r+1)! its r/(r+1)! . simple no ?:O


u need that for the HENCE bit.. which is what i cant do lol
that bits fine for me. its the other bit :)
i missed it out when i wrote problem but thats my fault -.-

and to that above me...
he ment multiply with r+1/r+1 because thats 1.. its a equal fraction so its ok :)

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 6:52 pm
by .AWAKE.
the answer is 4.

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 6:54 pm
by Toasty
the answer is: true.

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 6:57 pm
by penfold1992
.AWAKE. wrote:the answer is 4.

you forgot to carry the 1.

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 7:22 pm
by Blindfire
r = 1.

1/1-1/(1+1)=1/(1+1)

so 2/2 - 1/2 = 1/2.

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 7:22 pm
by Deadsolid
Permutations suck. Junior year was great though. Babs taught me...well, nothing i remember.

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 7:28 pm
by CeLL
Blindfire wrote:r = 1.

1/1-1/(1+1)=1/(1+1)

so 2/2 - 1/2 = 1/2.

thats surprisingly simple...

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 8:00 pm
by /Pi
1.1?

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 8:17 pm
by DumboDii
Blindfire wrote:r = 1.

1/1-1/(1+1)=1/(1+1)

so 2/2 - 1/2 = 1/2.


That's what I got too. But there could be more numbers which fulfil the formula so I wonder if solving one variable is enough.

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 9:38 pm
by penfold1992
DumboDii wrote:
Blindfire wrote:r = 1.

1/1-1/(1+1)=1/(1+1)

so 2/2 - 1/2 = 1/2.


That's what I got too. But there could be more numbers which fulfil the formula so I wonder if solving one variable is enough.


it works for all whole numbers above 0 at least. -.- it didnt prove it at all it just showed that it was true.

put 98363928692038694 as r and i bet you get the same outcome -.-

the reason is... because a PROOOF shows that. thats why the proof was necesarry before you do the 2nd part of the question.
proof by induction. kudos to any1 who actually does it rather then puts numbers into a formula already known to be true. also if it wasnt true why would it have a equal sign there?

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 9:52 pm
by KillAndChill
penfold1992 wrote:
DumboDii wrote:
Blindfire wrote:r = 1.

1/1-1/(1+1)=1/(1+1)

so 2/2 - 1/2 = 1/2.


That's what I got too. But there could be more numbers which fulfil the formula so I wonder if solving one variable is enough.


it works for all whole numbers above 0 at least. -.- it didnt prove it at all it just showed that it was true.

put 98363928692038694 as r and i bet you get the same outcome -.-

the reason is... because a PROOOF shows that. thats why the proof was necesarry before you do the 2nd part of the question.
proof by induction. kudos to any1 who actually does it rather then puts numbers into a formula already known to be true. also if it wasnt true why would it have a equal sign there?

Proof by induction is different than just proving an equation with variables. There are 2 main parts:
1)Prove the base case (the first case aka n=0)
2)Assuming the nth (any n) case is true, prove the (n+1)st case is also true

He proved it, just not by induction.

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 10:59 pm
by evilpeta
1/1! - 1/2! = 1/2!
1 - 1/2 = 1/2
1/2 = 1/2 => 1

1/R! - 1/(R+1)! = R/(R+1)!
(R+1)!/R! - (R+1)!/(R+1)! = R
(R+1) - 1 = R
R = R => 1

1/(R+1)! - 1/(R+2)! = (R+1)/(R+2)!
(R+2)!/(R+1)! - (R+2)!/(R+2)! = R+1
(R+2) - 1 = R+1
R+1 = R+1 => 1

-----------------------------------

∑ n/(n+1)! = ∑ [n+1-1/(n+1)!] = ∑ [(n+1)/(n+1)!] - ∑ [1/(n+1)!] = ∑ (1/n!) - ∑ [1/(n+1)!] =
∑ [1/n! - 1/(n+1)!] -> telescopic series

(1/1! - 1/2!) + (1/2! - 1/3!) + (1/3! - 1/4!) + ... + (1/n! - 1/(n+1)!) = .. (note: bold terms means they cancel)
1/1! - 1/(n+1)! = 1 - 1/(n+1)!

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 11:05 pm
by SidiousX
I fcking hate math.

Re: Math Problem 2)

Posted: Fri Oct 23, 2009 11:47 pm
by penfold1992
KillAndChill wrote:Proof by induction is different than just proving an equation with variables. There are 2 main parts:
1)Prove the base case (the first case aka n=0)
2)Assuming the nth (any n) case is true, prove the (n+1)st case is also true

He proved it, just not by induction.


hmm i see what you mean... but the sum of series between r=1 and n of
r/(r+1)! cannot be solved by proving it in the way he did.

seeing as you know what proof by induction is.. (unless u googled it haha)
wna give it a go :)... at least thats what i think needs to be done :S
summing series requires proof by induction? hmm cant be right... maybe summing of 1/x and such... **** FURTHER MATHS! WHY DID I TAKE IT!

Re: Math Problem 2)

Posted: Sat Oct 24, 2009 12:17 am
by Beethoven

Re: Math Problem 2)

Posted: Sat Oct 24, 2009 12:49 am
by KillAndChill
penfold1992 wrote:
KillAndChill wrote:Proof by induction is different than just proving an equation with variables. There are 2 main parts:
1)Prove the base case (the first case aka n=0)
2)Assuming the nth (any n) case is true, prove the (n+1)st case is also true

He proved it, just not by induction.


hmm i see what you mean... but the sum of series between r=1 and n of
r/(r+1)! cannot be solved by proving it in the way he did.

seeing as you know what proof by induction is.. (unless u googled it haha)
wna give it a go :)... at least thats what i think needs to be done :S
summing series requires proof by induction? hmm cant be right... maybe summing of 1/x and such... **** FURTHER MATHS! WHY DID I TAKE IT!

I would do it, but I hate it. It was a big part of my test today and I don't wanna ever do it again, lol.

Re: Math Problem 2)

Posted: Sat Oct 24, 2009 10:31 am
by penfold1992
hahaha this subject is absolute shite right? thats why i posted, i cant be asked with this stupid subject anymore xD

Re: Math Problem 2)

Posted: Sat Oct 24, 2009 12:14 pm
by evilpeta
i saw series in highschool and during integral calculus in the university, it was clearly the lighter and easier part of the subject lol. if you think summing series is shit, once you see diferential equations or multivariable calculus you will pull your ball hairs out! :banghead: