The Binomial Expansion

Edexcel IAL WMA12/01/P2/Jan 2020/Q2 (Binomial Expansion)

One of the terms in the binomial expansion of​​ 3 + ax6, where a is a constant, is​​ 540x4​​ 

• Find the possible values of a.​​ 

(4)​​ 

• Hence find the​​ term independent of x in the expansion of​​ 

181+1x63+ax6

(3)

SOLUTION​​ 

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

On expansion, the terms with​​ x4​​ appears to be the following.​​ 

n2an-4b4=6432ax4=540 x4

Hence, equating the coefficients.

15 x 9 x a4=540

135 a4=540

a4=540135

a4=4 

a4-4=0

Solving the quadratic​​ equation.​​ 

a2+2a2-2=0

a2+2=0      a2-2=0

a2=-2           a2-2=0

a=-2a2=2

      N.P                   a=±2

b- ​​ 

​​ The term independent of x is the one which is constant. So we need to find the constant terms in the expansion. One of​​ the constant term will form when two constant of the two bracket gets multiplied. Whereas the other is formed when​​ the​​ 1x6​​ is multiplied to the​​ x6​​ term of​​ 3+ax6​​ when binomially expanded.​​ 

When two constant of the two bracket gets multiplied.

181×36=9

When, the​​ 1x6​​ is multiplied to the​​ x6​​ term of​​ 3+ax6​​ when binomially expanded.​​ 

On the binomial expansion of​​ 3+ax6, the​​ x6​​ term we get is​​ 

n6an-6b6=6636ax6

=1×36×a6x6

=729a6x6

Now, when this​​ term gets multiplied with​​ 1x6, we get​​ 

=1x6×729a6x6

=729a6

Where,​​ a=±2​​ 

=729(±2)6

=2126

=23

=8

So, on combining both the terms that are independent of x, ​​ we get​​ 

=9+8=17