The Binomial Expansion

Coach Name: Sir Muhammad Abdullah Shah

Edexcel IAL WMA12/01/P2/Oct 2021/Q1 (Binomial Expansion)

The first three terms, in ascending powers of x, of the binomial expansion of​​ 1 + kx16​​ are​​ 1, -4x and px2.

where k and p are constants.​​ 

  • Find, in​​ simplest form,​​ 

  • the value of k​​ 

  • the value of p​​ 

(3)

gx=2+16x1+kx16

Using the value of k found in part (a),​​ 

  • find the term in x2​​ in the expansion of g(x).​​ 

(3)

SOLUTION​​ 

i-​​ Using binomial theorem to find the first 3 terms of the expression.​​ 

an=160116kx0=1 x 1 x 1=1

n1an-1 b=161115kx1=16 x 1 x kx=16 kx

n2an-3b3=162114-14x2=120 x 1 x116x2=152x2

Hence, the expansion​​ gives​​ 

1+ kx16=1+16kx+152x2+

Whereas it is given that the first three terms of the expansion are​​ 1, -4x and px2.​​ 

Lets equate the second term to find the value of k.​​ 

16kx= -4x

16k=-4

k=-416= --14

k=-14

ii- Equating the third term to find the value of p.​​ 

152x2=px2

Hence,​​ x2​​ gets cancelled out.​​ 

P=152

b- ​​ The expression we found with the​​ help of binomial expansion is​​ 

1+ kx16=1+16kx+152x2+

Lets substitute the value of​​ k=-14. Hence, the expression becomes​​ 

1-14x6=1-4x+152x2-354x3+.. 

Therefore,​​ 

gx=2+16x1+kx16

gx=2+16x1+ kx16=2+16x1-4x+152x2-354x3+.

Hence, the term in x2​​ in the expansion of g(x) are

2 x152x2+16x-354x3

15x2-140x2

-125x2