The Student Room Group
Reply 1
what've you tried so far?
Reply 2
well i know that (1+ax)^n will give

1 +(n)(ax) +(n)(n-1)(ax)^2 all over 2! + (n)(n-1)(n-2)(ax)^3 all over 3!

Then the question tells us that na = 8

and that (n)(n-1)(a^2)/2! = 30

stuck as i get answers which go into decimals, which is incorrect.

help!
Reply 3
So you have two silmultaneous equations, and two variables, so it can be solved. Rearrange the first to get a = n/8 and substitute into the second to find the values for n (remembering the actual n has to be greater than 2).
Reply 4
Lawbutwhere?
Hi
Cant figure this one out. any help GREATLY appreciated...

a) Write down the first 4 terms of the binomial expansion, in ascending powers of x, of (1+ax)^n, n >2.

b) Given that, in this expansion, the coefficient of x is 8 and the coefficient of x^2 is 30,

calculate the value of n and a.

c) Find the coefficient of x^3


a) (1 + ax)^n = 1 + anx + n(n - 1)a²x²/2! + n(n - 1)(n - 2)a³x³/3! + ...

b) an = 8, n(n - 1)a²/2 = 30

=> a = 8/n

n(n - 1)(8/n)² = 60
64(n - 1)/n = 60

64n - 64 = 60n

4n = 64

n = 16

a = 8/16 = 1/2

c) [16(16 - 1)(16 - 2)*(1/2)³]/3! = 2*15*14/6 = 70

=> Co-efficient of x^3 is 70.

Hope this helps,

~~Simba

Edit: Fixed a silly error :p: ...