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# P2 Expansion Question !!! watch

1. Hi people,

Is there any chance someone can help me solve the following problem. I have tried it several times and I keep getting into a serious mess!!!

Q) In the expansion of (1+x/2)^n the coefficients of the terms in x^2 and x^3 are equal. Find n.

I have tried to solve it by putting the formulas for finding the coefficients in x^2 and x^3 equal to eachother (duhh) but I just get completely screwed by the result. I know that n=8 but I can't prove it!!!

PS - there is rep at stake here!!
2. (Original post by Leekey)
Hi people,

Is there any chance someone can help me solve the following problem. I have tried it several times and I keep getting into a serious mess!!!

Q) In the expansion of (1+x/2)^n the coefficients of the terms in x^2 and x^3 are equal. Find n.

I have tried to solve it by putting the formulas for finding the coefficients in x^2 and x^3 equal to eachother (duhh) but I just get completely screwed by the result. I know that n=8 but I can't prove it!!!

PS - there is rep at stake here!!

Rep at stake, here I come

(1+x/2)^n = 1 + nx/2 + n(n-1)/2 . (x/2)^2 + n(n-1)(n-2)/3!.(x/2)^3

Therefore, n(n-1)/8 = n(n-1)(n-2)/48 with n > 1 I assume, otherwise it's trivial. So n-2 = 6, so n=8.
3. (Original post by Leekey)
Hi people,

Is there any chance someone can help me solve the following problem. I have tried it several times and I keep getting into a serious mess!!!

Q) In the expansion of (1+x/2)^n the coefficients of the terms in x^2 and x^3 are equal. Find n.

I have tried to solve it by putting the formulas for finding the coefficients in x^2 and x^3 equal to eachother (duhh) but I just get completely screwed by the result. I know that n=8 but I can't prove it!!!

PS - there is rep at stake here!!
(1+x/2)^n = 1 + nx/2 + n(n-1)/2 x^2/4 + n(n-1)(n-2)/6 x^3 /8 + ...
n(n-1)/8 = n(n-1)(n-2)/48

1 = (n-2)/6

n-2 = 6

n=8
4. (Original post by theone)
Rep at stake, here I come

(1+x/2)^n = 1 + nx/2 + n(n-1)/2 . (x/2)^2 + n(n-1)(n-2)/3!.(x/2)^3

Therefore, n(n-1)/8 = n(n-1)(n-2)/48 with n > 1 I assume, otherwise it's trivial. So n-6 = 3, so n=9.
but n=8
5. (Original post by elpaw)
but n=8
Check it
6. (Original post by theone)
Check it
even leekey said n=8

where did you get n-6=3 from?
7. (Original post by elpaw)
even leekey said n=8

where did you get n-6=3 from?
typo.
8. (Original post by theone)
Rep at stake, here I come

(1+x/2)^n = 1 + nx/2 + n(n-1)/2 . (x/2)^2 + n(n-1)(n-2)/3!.(x/2)^3

Therefore, n(n-1)/8 = n(n-1)(n-2)/48 with n > 1 I assume, otherwise it's trivial. So n-2 = 6, so n=8.
WAA??
I'm doing P2 and you've just lost me here mate..lol
9. (Original post by jndikum)
WAA??
I'm doing P2 and you've just lost me here mate..lol
n>1 because otherwise you don't get a coeff of x^2 or x^3....and apart from that it's just binomial expansion...you'll come to it
10. (Original post by king of swords)
n>1 because otherwise you don't get a coeff of x^2 or x^3....and apart from that it's just binomial expansion...you'll come to it
I'm on the OCR syllabus..which are you on?
11. (Original post by theone)
typo.
crafty editing there. trying to win that rep, are you?
12. (Original post by jndikum)
I'm on the OCR syllabus..which are you on?
edexcel...remember, the evil examining board who made a mess up on a maths exam a few years back .
13. (Original post by elpaw)
crafty editing there. trying to win that rep, are you?
HAHAHA!! That's what I noticed.
14. (Original post by elpaw)
crafty editing there. trying to win that rep, are you?
Won't say no... and to be perfectly honest, I edited it straight after I posted it...
15. (Original post by theone)
Won't say no... and to be perfectly honest, I edited it straight after I posted it...
You and elpaw type too quick for me...i was going to get that rep .
16. (Original post by theone)
Won't say no... and to be perfectly honest, I edited it straight after I posted it...
straight after i pointed it out
17. (Original post by Adhsur)
HAHAHA!! That's what I noticed.
How dare you
18. so where is leekey to reap the rewards
19. (Original post by elpaw)
so where is leekey to reap the rewards
i think he tricked us all
20. (Original post by king of swords)
i think he tricked us all
as long as he doesnt come over all maskallesque and say he "already knew the answer hahahahahaha"

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