The Binomial Expansion

Edexcel IAL WMA12/01/P2/Jan 2022/Q3 (Binomial Expansion)

• Find the first 4 terms, in ascending powers of x, of the binomial expansion of​​ 

2-kx48

where k is a non‑zero constant. Give each term​​ in simplest form.​​ 

(4)​​ 

fx=5-3x2-kx48

In the expansion of f(x), the constant term is 3 times the coefficient of x.​​ 

• Find the value of k.​​ 

(3)

SOLUTION​​ 

a-​​ Using binomial theorem to find the first 4 terms of the expression.​​ 

Finding the first four terms separately and will then arrange them in the ascending order of power of​​ x.​​ 

an=8028-kx40=1 ×256×1=256

n1an-1 b=8127-kx41= 8 ×128×-kx4=-256 kx

n2an-2b2=8226-kx42=28×64×k2x216=112 k2x2

n3an-3b3=8325-kx43= 56×32×-k3x364

2-kx48=256-256 kx+112k2x2-28k3x3

b-​​ 

fx=5-3x2-kx48

fx=5-3x256-256 kx+112k2x2-28k3x3

fx=5256+5-256kx+5112k2x2+5-28k3x3-3x256-3x-256kx…

Not expanding further as on multiplying we will get x with a power greater than 1. Which is not required rightnow.​​ 

It is given in the question that the constant term is 3 times the​​ coefficient of x.​​ 

Where, on expansion of f(x), constant is​​ 5 x 256

And the coefficient of x is​​ 5-256k-3256

So, according to the statement of the question

constant=3×coefficient of x

5 x 256=3×5-256k-3256

1280=3-1280 k-768

1280=-3840 k-2304

3840k=-2304-1280

3840k=-3584

k=-35843840.

k=-1415