C2 Binomial

I worked through most of the Questions in the Binomial chapter of the Green C1 and C2 book by David Rayner and Paul williams and did all the question sucessfully except for this question pg 164, Q25

If anyone helps me with this i will be very grateful

f(x)= (1+x/k)^n

Given that the coefficient of x^3 is twice the coefficient of x^2 in the binomial expansion of f(x)

a) Prove that n = 6k+2

Given also that the coefficients of x^4 and x^5 are equal and non-zero,

b)form another equation in n and k and hence show that k = 2 and n = 14

Using these values of k and n,

c)expand f(x) in ascending powers of x, up to and including the term in x^5.
Give each coefficient as an exact fraction in its lowest terms

Thanks if anyone can help :smile:

Reply 1

yusufu

dec0!

I worked through most of the Questions in the Binomial chapter of the Green C1 and C2 book by David Rayner and Paul williams and did all the question sucessfully except for this question pg 164, Q25

If anyone helps me with this i will be very grateful

f(x)= (1+x/k)^n

Given that the coefficient of x^3 is twice the coefficient of x^2 in the binomial expansion of f(x)

a) Prove that n = 6k+2
Coeff of x^3 = 1/6[n(n-1)(n-2)/k³]
Coeff of x^2 = 1/2[n(n-1)/k²]

1/6[n(n-1)(n-2)/k³] = 2* 1/2[n(n-1)/k²]
Cancelling on both sides gives:
(n-2)/(6k) = 1
n = 6k+2

Reply 2

yusufu

Given also that the coefficients of x^4 and x^5 are equal and non-zero,

b)form another equation in n and k and hence show that k = 2 and n = 14

Coeff x^4 = 1/24 [n(n-1)(n-2)(n-3)/k⁴
Coeff x^5 = 1/120 [n(n-1)(n-2)(n-3)(n-4)/k⁵

1/120 [n(n-1)(n-2)(n-3)(n-4)/k⁵ = 1/24 [n(n-1)(n-2)(n-3)/k⁴
Cancelling gives:
(n-4)/(120k) = 1/24
n = 5k+4

From previous:
n = 6k+2 = 5k+4
k = 2
n=14

Reply 3

yusufu

Using these values of k and n,

c)expand f(x) in ascending powers of x, up to and including the term in x^5.
Give each coefficient as an exact fraction in its lowest terms
= 1 + 14x + (1/2)(1/4)(14.13 x^2) + (1/6)(1/8)(14.13.12 x^3) + (1/24)(1/16)(14.13.12.11 x^4) + (1/120)(1/32)(14.13.12.11.10 x^5)

I can't be bothered to simplify it so I'll let you do that.