Binomial Theorem Question

If the coefficients of x^4 and x^6 of expansion (1+2x)^n are in the ratio 5:8, find n.

Would appreciate working thanks! :)

Those are the binomial coefficients times some power of two. Alright, I'm gonna write the binomial coefficient as C(n,k). That is:

C(n,k) = n! / ( k! (n-k)! ) where the ! means the factorial, Eg. 5! = (5)(4)(3)(2)(1)

Thus, by the Binomial Theorem, the 4th term when expanding your equation must be (to simplify the counting, the very first term I'm taking it to be the 0th term):

C(n,4) 1^(n-4) (2x)^4 = 16 C(n,4) x^4
The 6th term is

C(n,6) 1^(n-6) (2x)^6 = 64 C(n,6) x^6

Since the ratio is 5:8, it means that 16 C(n,4) / ( 64 C(n,6) ) = 5 / 8
Therefor

2 C(n,4) = 5 C(n,6)

With a little bit of algebra,

2 n! / ( 4! (n-4)! ) = 5 n! / (6! (n-6)!)
Cancelling terms, and reducing the equation, we get

12 (n-6)! = (n-4)! But since (n-4)! = (n-4) (n-5) (n-6)! we can reduce this further more... Then

12 = (n-4)(n-5)

12 = n^2 -9n +20
or
n^2 -9n +8=0
Solving for n we get

n=1 or n=8.

Obviously n=1 can't be the right answer since you want that Binomial to be elevated to at least the 6th power, ergo the only possible answer is n=8.

fifth term will have x^4 and 7 th term will have x^6

coefficient of x^4 = nC4 2^4 = 16*nC4

coefficient of x^6 = nC6 2^6 = 64*nC6

==> (nC4/nC6)(16/64) = 5/8

nC4/nC6 = (64*5)/(16*8) = 5/2

n (n-1)(n-2)(n-3) /n(n-1)(n-2)(n-3)(n-4)(n-5) * (6!/4!) = 5/2

30/(n-4)(n-5) = 5/2

6/(n^2 - 9n + 20) = 1/2

n^2 - 9n + 20 = 12

=> n^2 - 9n + 8 = 0

=>(n - 8)(n - 1) = 0

valid solution is n = 8