Sequences & Series

Edexcel IAL WMA12/01/P2/Jan 2023/Q3 (Binomial Expansion, Arithmetic Series)

fx=2+kx8' where k is a non‑zero constant

• Find the first 4 terms, in ascending powers of x, of the binomial expansion of f (x). Give each term in simplest form.​​ 

(4)​​ 

Given that, in the binomial expansion of f (x), the coefficients of x, x2​​ and x3​​ are the first 3 terms of an arithmetic progression,​​ 

• find, using algebra, the possible values of k.​​ 

(Solutions relying entirely on calculator technology are not acceptable.)​​ 

(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=70 27 kx80=1 x 128 x 1=128

n1 an-1 b=71 26  kx81 =7 x 64 xkx8=56 kx 

n2 an-2 b2=72 25  kx82=21 x 32 xk2x264=212k2x2

n3 an-3 b3=73 24  kx83=35 x 16 xk3 x3512 =3523k3 x3 

2+kx87=128+56kx+212k2x2+3532 k3x3  

b- To find the possible values of k, taking help of the information that the coefficient of​​ x, x2​​ and​​ x3​​ are the first three terms of the arithmatic sequence.​​ 

(Remember, in an arithmetic sequence, difference between the consecutive terms is constant. So equating the difference between the 1st​​ and 2nd​​ term to the difference between 2nd​​ and 3rd​​ term gives us the polynomial equation of k that we can solve and find the possible values of k.) ​​ 

So the arithmetic sequence is​​ 

56k, 212k2 , 3532 k3, … 

difference between 3rd and 2nd term=difference between 2nd and 1st term

3532 k3-212 k2=212k2-56k

3532 k3-21k2+56k=0   ÷7

532 k3-3k2+8k=0

5k3-96k2+256k=0 

k5k2- 96k+256= 0

5k2-96k+256=0

a=5        b= -96        c=256 

k=--96±-962-4552625 

k= 16        ,     k=165