[aeyurttaser13]

hey guys! i think this was by far the most difficult question in paper 1 yesterday, the only one i couldn't reach an answer in.. if u guys know how to solve it please can u help me learn the way to solve it? ive never seen a question so complicating in binomial theorem.

if u dont remember, it was:
(1+2x/3)^n x (3+nx)^2 = 9 + 84x ....
find n?

[Iaintinkler]

Expanding the second bracket gave 6xn + 9 +... The only value from the 1st bracket that could be used was the x^1 regardless of what n was, and so 2/3 of 9 is 6. I then took this away from 84 to give 78 and divided by 6 to give the value which 6x must be multiplied by to give 78 = 13.
Unfortunately, I forgot to include the 2/3 of 9 until after the exam

[aeyurttaser13]

(Original post by Iaintinkler)
Expanding the second bracket gave 6xn + 9 +... The only value from the 1st bracket that could be used was the x^1 regardless of what n was, and so 2/3 of 9 is 6. I then took this away from 84 to give 78 and divided by 6 to give the value which 6x must be multiplied by to give 78 = 13.
Unfortunately, I forgot to include the 2/3 of 9 until after the exam

ooh sorry to hear that.. i did expand the 2nd bracket and all.. but i never thought of using logic to solve it cz since it was a noncalculator paper i thot using logic or trial and error would not get me more than a mark.. then when i went back to it i was running outta time so i just showed some working but i dont think ill get over 2-3 points. what was the rest of the paper like? whatr u expecting?

[chicheto93]

I expanded the second bracket and insterted the 1 in afterwards and I got 14. But a lot of people are saying the answer is 7. Does anyone know for sure the correct answer?

[dstandish]

Hey

Yay i got n=13 I had to go back to it at the end though. Tried logs and everything just didn't really know how to approach it. But the easiest way to solve it is:

(1+2/3x)^n(3+nx)^2=9+84x

Expand (3+nx)^2 you get: 9+6nx+n^2x^2
So use 6nx and the coefficient from the other bracket which is 1^n. 1 to the power of anything is 1 so you know you have 6nx.

For the first bracket you have 2/3x times by 9 (the coefficient from the second pair of brackets (see above where I expanded.)

So you have 6x + 6nx = 84 x and then solve.

I guess its one that looks so hard at first... really didn't think I was gonna get an answer.. but pretty easy when you know how.

[dstandish]

Reading the last post maybe I got it wrong.... but pretty certain thats the answer

[dstandish]

ah well at least we get method marks

[aeyurttaser13]

(Original post by dstandish)
Hey

Yay i got n=13 I had to go back to it at the end though. Tried logs and everything just didn't really know how to approach it. But the easiest way to solve it is:

(1+2/3x)^n(3+nx)^2=9+84x

Expand (3+nx)^2 you get: 9+6nx+n^2x^2
So use 6nx and the coefficient from the other bracket which is 1^n. 1 to the power of anything is 1 so you know you have 6nx.

For the first bracket you have 2/3x times by 9 (the coefficient from the second pair of brackets (see above where I expanded.)

So you have 6x + 6nx = 84 x and then solve.

I guess its one that looks so hard at first... really didn't think I was gonna get an answer.. but pretty easy when you know how.

damn it now it seems so clear! i wish i had more than 5 mins after being done with the exam.. cz i kept on double checking each q im done with so that i dont make silly mistakes. ugh wish i hadnt done that id have more than 5 mins for q 7

[dstandish]

Yeah... i think its because it looks like binomial... but you don't really need binomial at all to solve it. IB are such mind-****s

[chicheto93]

(Original post by dstandish)
Hey

Yay i got n=13 I had to go back to it at the end though. Tried logs and everything just didn't really know how to approach it. But the easiest way to solve it is:

(1+2/3x)^n(3+nx)^2=9+84x

Expand (3+nx)^2 you get: 9+6nx+n^2x^2
So use 6nx and the coefficient from the other bracket which is 1^n. 1 to the power of anything is 1 so you know you have 6nx.

For the first bracket you have 2/3x times by 9 (the coefficient from the second pair of brackets (see above where I expanded.)

So you have 6x + 6nx = 84 x and then solve.

I guess its one that looks so hard at first... really didn't think I was gonna get an answer.. but pretty easy when you know how.

I am quite sure that this wrong because:

(1+2x/3)^n * (3+xn)^2
Expand the second term, now we have (1+2x/3)^n * (9+6xn+(xn)^2)
So we know that the coefficient of x in the expansion is 84
The only way we can get x is by multiplying the (1) from the first bracket by (6xn) from the second bracket and (2x/3) from the first bracket by (9) from the second bracket.
However, the 2x/3 term is the SECOND term of the binomial expansion, which means that, according to the theorem, you need to multiply the term by (n!)/(n-1)!, which is simply (n).
So now we have (2xn/3)*(9)+(6xn)(1)=84x
6xn+6xn=84x
n=7

It's either this or 14 ^^ but I think this seems like a more reasonable solution.

[Saymyname!]

Higher math peep here. Interesting question! I've attached the working out--> n=7. In a way that will get you all annoyed... the answer to question 7... is 7!

[aeyurttaser13]

(Original post by Saymyname!)
Higher math peep here. Interesting question! I've attached the working out--> n=7. In a way that will get you all annoyed... the answer to question 7... is 7!

thx a lot.. now it seems much more logical. damn the only q i missed

[Ulmus Parvifolia]

I used binomial expansion, got n = 14. I may have made a mistake while calculating, though. Anyway, my final equation was 6 nx = 84 x

[doctor92]

I just forgot to add (n nCr 0) part at the beginning how many points would I lose ? anyone have any idea this is so annoying now..

[Kevjung]

(Original post by doctor92)
I just forgot to add (n nCr 0) part at the beginning how many points would I lose ? anyone have any idea this is so annoying now..

I couldn't solve this but isn't that 1 anyways?

[doctor92]

(Original post by Kevjung)
I couldn't solve this but isn't that 1 anyways?

yes it is, but then you have to multiply it with the other bracket to find the final equation for x I just can't believe I have lost 5 points or more because of that.

[aeyurttaser13]

(Original post by doctor92)
yes it is, but then you have to multiply it with the other bracket to find the final equation for x I just can't believe I have lost 5 points or more because of that.

doesn't the IB apply "error carried forward"? like u should at least get marks if u are on track

[doctor92]

(Original post by aeyurttaser13)
doesn't the IB apply "error carried forward"? like u should at least get marks if u are on track

I don't know I hope so.. I have checked the mark scheme for a similar question in 2010, they only gave 2 points for method, for a) attempting to expand what's in the brackets and b) attempting to form an equation at the end for x... But this is a 7 points question.. maybe they give more
I'm just hoping it's my only silly mistake

[aeyurttaser13]

(Original post by doctor92)
I don't know I hope so.. I have checked the mark scheme for a similar question in 2010, they only gave 2 points for method, for a) attempting to expand what's in the brackets and b) attempting to form an equation at the end for x... But this is a 7 points question.. maybe they give more
I'm just hoping it's my only silly mistake

owh :/ im really hoping the same, i dont want to lose more marks after losing abt 5 points on that q as well

[Roudin100]

it turned out to be 7, but you guys shouldn't worry, I mean you've got about 15 marks to lose in each paper and still be in the boundaries of a seven. Chill and focus on your next exam (: