C2 HEELP

[T.Q]

The first four terms, in ascending powers of x, of the binomial expansion of (1 + kx)^n are

1 + Ax + Bx2 + Bx3 + …,

where k is a positive constant and A, B and n are positive integers.

(a) By considering the coefficients of x2 and x3, show that 3 = (n – 2) k.    (4 MARKS)

Given that A = 4,

(b) find the value of n and the value of k.     (4 MARKS)



PLZ SHOW YOUR WORKING

[T.Q]

heeeeeeeeeelp
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[astarmathsandphysics]

trythis.

[XFlufferzX]

Incase if my handwriting wasnt clear there ya go :
Using the identity:
(1+kx)^n = 

1 + n (kx) + n (kx) + n(n-1)/2! k^2 + n(n-1)(n-2)/3! kx^3

1+ Ax + Bx^2 + Bx^3

so A = nk
    B=  n(n-1)/2! k^2
    B= n(n-1)(n-2)/3! k^3

(2!= 2 , 3! =6)

Equating the problem:

n(n-1)/2 k^2 = n(n-1)(n-2)/6 k^3

Simplify it and you'll get

3= (n-2) k

For b)

3 = (n-2) k
Simplify

3 = nk -2k

3 = A -2k

3 = 4 - 2k

K = 1/2

Solving for n

A = nk

4 = n(0.5)

n=8

Its a good question, from which paper did you get dat?

[T.Q]

thank you both

its from the mock paper

[T.Q]

Incase if my handwriting wasnt clear there ya go :
Using the identity:
(1+kx)^n = 

1 + n (kx) + n (kx) + n(n-1)/2! k^2 + n(n-1)(n-2)/3! kx^3

1+ Ax + Bx^2 + Bx^3

so A = nk
    B=  n(n-1)/2! k^2
    B= n(n-1)(n-2)/3! k^3

(2!= 2 , 3! =6)

Equating the problem:

n(n-1)/2 k^2 = n(n-1)(n-2)/6 k^3

Simplify it and you'll get

3= (n-2) k

For b)

3 = (n-2) k
Simplify

3 = nk -2k

3 = A -2k

3 = 4 - 2k

K = 1/2

Solving for n

A = nk

4 = n(0.5)

n=8

Its a good question, from which paper did you get dat?

the paper is attached

[XFlufferzX]

thanks!