Solution of Past Paper of CAIE Math 9709/12 Feb/Mar 2020 Question # 6

CIE/GCE/AS/Math/P1/20/Mar/12/Q#6

Question

1 𝑎 5
The coefficient of in the expansion of (2𝑥 + 2
) is 720.
𝑥 𝑥



a) Find the possible values of the constant 𝑎.
1
b) Hence find the coefficient of 1 in the expansion.
𝑥7


Solution

a)


We are given expression as;
𝑎 5
(2𝑥 + )
𝑥2
Expression for the general term in the Binomial expansion of (𝑥 + 𝑦)𝑛 is:


𝑛
𝑇𝑟+1 = ( ) 𝑥 𝑛−𝑟 𝑦 𝑟
𝑟


In the given case:
𝑎
𝑥 = 2𝑥 , 𝑦= , 𝑛=5
𝑥2
Hence;
𝑛
𝑇𝑟+1 = ( ) 𝑥 𝑛−𝑟 𝑦 𝑟
𝑟


5 𝑎 𝑟
𝑇𝑟 +1 = ( ) (2𝑥)5−𝑟 ( 2 )
𝑟 𝑥

𝑟
5 ( )5−𝑟 ( )5−𝑟 ( )𝑟 ( 1 )
𝑇𝑟 +1 =( ) 2 𝑥 𝑎
𝑟 𝑥2

, CIE/GCE/AS/Math/P1/20/Mar/12/Q#6


5
𝑇𝑟 +1 = ( ) (2)5−𝑟 (𝑥 )5−𝑟 (𝑎)𝑟 (𝑥 −2 )𝑟
𝑟


5
𝑇𝑟+1 = ( ) (2)5−𝑟 (𝑥 )5−𝑟 (𝑎)𝑟 (𝑥 )−2𝑟
𝑟


5
𝑇𝑟 +1 = ( ) (2)5−𝑟 (𝑥 )5−𝑟−2𝑟 (𝑎)𝑟
𝑟


5
𝑇𝑟+1 = ( ) (2)5−𝑟 (𝑥 )5−3𝑟 (𝑎)𝑟
𝑟

1
Since we are looking for the coefficient of : we can equate;
𝑥



1
(𝑥 )5−3𝑟 =
𝑥
(𝑥 )5−3𝑟 = 𝑥 −1
5 − 3𝑟 = −1
3𝑟 = 6
𝑟=2


1
Now we can find the term with ; Substituting 𝑟 = 2;
𝑥



5
𝑇2 +1 = ( ) (2)5−2 (𝑥 )5−3(2) (𝑎)2
2


5
𝑇3 = ( ) (2)3 (𝑥 )5−6 (𝑎)2
2