CIE/GCE/AS/Math/P1/19/Nov/13/Q#1
Question
(i) Expand (1 + 𝑦)6 in ascending powers of y as far as the term in y 2.
6
(ii) In the expansion of (1 + (𝑝𝑥 − 2𝑥 2 )) the coefficient of x 2 is 48. Find the
value of the positive constant 𝑝.
Solution
i.
We are required to expand (1 + 𝑦)6.
Expression for the Binomial expansion of (𝑥 + 𝑦)𝑛 is:
𝑛
𝑛
(𝑥 + 𝑦)𝑛 = ∑ ( ) 𝑥 𝑛−𝑟 𝑦 𝑟
𝑟
𝑟=0
𝑛 𝑛 𝑛 𝑛 𝑛
(𝑥 + 𝑦)𝑛 = ( ) 𝑥 𝑛 𝑦 0 + ( ) 𝑥 𝑛−1 𝑦1 + ( ) 𝑥 𝑛−2 𝑦 2 + ⋯ … + ( ) 𝑥 1 𝑦𝑛−1 + ( ) 𝑥 0 𝑦 𝑛
0 1 2 𝑛−1 𝑛
In the given case:
𝑥 =1, 𝑦=𝑦, 𝑛=6
Hence;
(1 + 𝑦)6 = (6) (1)6 (𝑦)0 + (6) (1)5(𝑦)1 + (6) (1)4 (𝑦)2 + (6) (1)3 (𝑦)3 … …
0 1 2 3
6! 6! 6!
(1 + 𝑦 )6 = (1 )5 ( 𝑦 )0 + (1)4 (𝑦)1 + (1)3 (𝑦)2
(6 − 0)! × 0! (6 − 1)! × 1! (6 − 2)! × 2!
6!
+ (1)3 (𝑦)3 + ⋯
(6 − 3)! × 2!
6! 6! 6! 6!
(1 + 𝑦 )6 = (1)(1) + (1)(𝑦) + (1)(𝑦)2 + (1)(𝑦)3 + ⋯
6! × 0! 5! × 1! 4! × 2! 3! × 3!
Question
(i) Expand (1 + 𝑦)6 in ascending powers of y as far as the term in y 2.
6
(ii) In the expansion of (1 + (𝑝𝑥 − 2𝑥 2 )) the coefficient of x 2 is 48. Find the
value of the positive constant 𝑝.
Solution
i.
We are required to expand (1 + 𝑦)6.
Expression for the Binomial expansion of (𝑥 + 𝑦)𝑛 is:
𝑛
𝑛
(𝑥 + 𝑦)𝑛 = ∑ ( ) 𝑥 𝑛−𝑟 𝑦 𝑟
𝑟
𝑟=0
𝑛 𝑛 𝑛 𝑛 𝑛
(𝑥 + 𝑦)𝑛 = ( ) 𝑥 𝑛 𝑦 0 + ( ) 𝑥 𝑛−1 𝑦1 + ( ) 𝑥 𝑛−2 𝑦 2 + ⋯ … + ( ) 𝑥 1 𝑦𝑛−1 + ( ) 𝑥 0 𝑦 𝑛
0 1 2 𝑛−1 𝑛
In the given case:
𝑥 =1, 𝑦=𝑦, 𝑛=6
Hence;
(1 + 𝑦)6 = (6) (1)6 (𝑦)0 + (6) (1)5(𝑦)1 + (6) (1)4 (𝑦)2 + (6) (1)3 (𝑦)3 … …
0 1 2 3
6! 6! 6!
(1 + 𝑦 )6 = (1 )5 ( 𝑦 )0 + (1)4 (𝑦)1 + (1)3 (𝑦)2
(6 − 0)! × 0! (6 − 1)! × 1! (6 − 2)! × 2!
6!
+ (1)3 (𝑦)3 + ⋯
(6 − 3)! × 2!
6! 6! 6! 6!
(1 + 𝑦 )6 = (1)(1) + (1)(𝑦) + (1)(𝑦)2 + (1)(𝑦)3 + ⋯
6! × 0! 5! × 1! 4! × 2! 3! × 3!
