# P2 Expansion Question !!!

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#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!!
0
16 years ago
#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.
0
16 years ago
#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
0
16 years ago
#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
0
16 years ago
#5
(Original post by elpaw)
but n=8
Check it 0
16 years ago
#6
(Original post by theone)
Check it even leekey said n=8

where did you get n-6=3 from?
0
16 years ago
#7
(Original post by elpaw)
even leekey said n=8

where did you get n-6=3 from?
typo.
0
16 years ago
#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
0
16 years ago
#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 0
16 years ago
#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?
0
16 years ago
#11
(Original post by theone)
typo.
crafty editing there. trying to win that rep, are you?
0
16 years ago
#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 .
0
16 years ago
#13
(Original post by elpaw)
crafty editing there. trying to win that rep, are you?
HAHAHA!! That's what I noticed. 0
16 years ago
#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...
0
16 years ago
#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 .
0
16 years ago
#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
0
16 years ago
#17
HAHAHA!! That's what I noticed. How dare you 0
16 years ago
#18
so where is leekey to reap the rewards
0
16 years ago
#19
(Original post by elpaw)
so where is leekey to reap the rewards
i think he tricked us all 0
16 years ago
#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"
0
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