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# Yet anorher p2 Question.... Binomial Expansion watch

1. The coefficient of x^2 in the expansion of (1+3x)^n is 252
Given that n is a positive integer,
a) Find the value of n
b) show that the coefficient of x^3 is 1512

Once I know the value of n, part b) will be a piece of cake. However, I haven't got the blindest idea of how to go about solving part a).
2. (Original post by Les Paul)
The coefficient of x^2 in the expansion of (1+3x)^n is 252
Given that n is a positive integer,
a) Find the value of n
b) show that the coefficient of x^3 is 1512

Once I know the value of n, part b) will be a piece of cake. However, I haven't got the blindest idea of how to go about solving part a).
Well if you expand (a+x)^n, you get:

a^n + a^(n-1)x + a^(n-2)x^2 + a^(n-3)x^3 .... a^2x^(n-2) + ax^(n-1) + x^n, with coefficients dependant on pascal's triangle.

Look at the x^2 one: a^(n-2)x^2
It is the third from the left, so if you look up on pascal's traingle the row which has 252 as the third number in from the left, that row should be the value of n. But then you have 3x, not just x. Which just completely confused me.

I don't know.. do you have a formula, I think there is one involving factorals. Maybe just substituting values in will give you an answer.
3. (Original post by Les Paul)
The coefficient of x^2 in the expansion of (1+3x)^n is 252
Given that n is a positive integer,
a) Find the value of n
b) show that the coefficient of x^3 is 1512

Once I know the value of n, part b) will be a piece of cake. However, I haven't got the blindest idea of how to go about solving part a).
well (1+3x)^2=1 + n*3nx+n(n-1)9x^2/2

so 9n(n-1)/2=252
so 9n(n-1)=504

so n(n-1)=56 so n=8
4. (Original post by mik1a)
Well if you expand (a+x)^n, you get:

a^n + a^(n-1)x + a^(n-2)x^2 + a^(n-3)x^3 .... a^2x^(n-2) + ax^(n-1) + x^n, with coefficients dependant on pascal's triangle.

Look at the x^2 one: a^(n-2)x^2
It is the third from the left, so if you look up on pascal's traingle the row which has 252 as the third number in from the left, that row should be the value of n. But then you have 3x, not just x. Which just completely confused me.

I don't know.. do you have a formula, I think there is one involving factorals. Maybe just substituting values in will give you an answer.
yeh u get (1+x)^n

u get 1 + n!x/(n-1)! [which equals n] +n!x^2/(n-2)!2! [which equals n(n-1)/2]
+ n!x^3/(n-3)!3! [which equals n(n-1)(n-2)/6
5. Ok I'm sorry, I'm still not really clear on this. I don't really understand all this factorial stuff.
6. (Original post by Les Paul)
Ok I'm sorry, I'm still not really clear on this. I don't really understand all this factorial stuff.
factorial means just times everything by everything below it
so, 5! is 5x4x3x2x1
2! is 2x1 etc etc

you can use this to work out these binomials.

ive attached a formula. it ONLY works for (1+K)^n
ive put kx = K but it can be anything.

if you open out your question to the x^2 term, youll be able to equate and find that n = 8.
7. (Original post by kikzen)
factorial means just times everything by everything below it
so, 5! is 5x4x3x2x1
2! is 2x1 etc etc

you can use this to work out these binomials.

ive attached a formula. it ONLY works for (1+K)^n
ive put kx = K but it can be anything.

if you open out your question to the x^2 term, youll be able to equate and find that n = 8.
Thanks a lot man. That cleared it up quite a bit.

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