1. Further binomial
Series
expansion

2. Series
KUS objectives
BAT complete problems involving partial fractions followed by series expansion
Starter:
Write 𝑥 −2 , − 1 2 <𝑥< in ascending powers of x up to the term in 𝑥 2
Hence, find the expansion of 𝑥− 𝑥 up to the term in 𝑥 2
1+2𝑥 −2 =1 − 4𝑥+12 𝑥 2 +…
𝑥− 𝑥 = −3 + 13𝑥 − 40 𝑥 2 + …

3. Write 4−5𝑥 (1+𝑥)(2−𝑥) in partial fractions
WB10a matching coefficients
Write 4−5𝑥 (1+𝑥)(2−𝑥) in partial fractions
Hence, find the expansion of 4−5𝑥 (1+𝑥)(2−𝑥) up to and including the term in 𝑥 3
4−5𝑥 (1+𝑥)(2−𝑥) = 𝐴 1+𝑥 + 𝐵 2−𝑥
4−5𝑥 = 𝐴 2−𝑥 + 𝐵(1+𝑥)
𝐴= 𝐵=−2
4−5𝑥 (1+𝑥)(2−𝑥) = 3 1+𝑥 − 2 2−𝑥
Check by adding together!
4−5𝑥 (1+𝑥)(2−𝑥) = 3 1+𝑥 −1 −2 2−𝑥 −1

4. 4−5𝑥 (1+𝑥)(2−𝑥) =3 1−𝑥+ 𝑥 2 − 𝑥 3 + … − 1+ 1 2 𝑥+ 1 4 𝑥 2 + 1 8 𝑥 3 +…
WB 10b
b) Hence, find the expansion of 4−5𝑥 (1+𝑥)(2−𝑥) up to and including the term in 𝑥 3
𝟏+𝒂𝒙 𝒏 =𝟏+𝒏 𝒂𝒙 + 𝒏 𝒏−𝟏 𝟐! 𝒂𝒙 𝟐 + 𝒏 𝒏−𝟏 (𝒏−𝟐) 𝟑! 𝒂𝒙 𝟑 + …
4−5𝑥 (1+𝑥)(2−𝑥) = 3 1+𝑥 −1 −2 2−𝑥 −1
3 1+𝑥 −1 = −𝑥+ 𝑥 2 − 𝑥 3 + …
2 2−𝑥 −1 = 1− 1 2 𝑥 −1 =
𝑥 𝑥 𝑥 3 +…
4−5𝑥 (1+𝑥)(2−𝑥) =3 1−𝑥+ 𝑥 2 − 𝑥 3 + … − 𝑥 𝑥 𝑥 3 +…
4−5𝑥 (1+𝑥)(2−𝑥) =2− 7 2 𝑥 𝑥 2 − 𝑥 3 +…

5. Write 2𝑥+7 (𝑥+3)(𝑥+4) in partial fractions
WB11a a matching coefficients
Write 2𝑥+7 (𝑥+3)(𝑥+4) in partial fractions
Hence, find the expansion of 2𝑥+7 (𝑥+3)(𝑥+4) up to and including the term in 𝑥 3
2𝑥+7 (𝑥+3)(𝑥+4) = 𝐴 𝑥+3 + 𝐵 𝑥+4
2𝑥+7 = 𝐴 𝑥+4 + 𝐵(𝑥+3)
𝐴= 𝐵=1
2𝑥+7 (𝑥+3)(𝑥+4) = 1 𝑥 𝑥+4
Check by adding together!
2𝑥+7 (𝑥+3)(𝑥+4) = 𝑥+4 −1 + 𝑥+3 −1
= 4+𝑥 − 𝑥 −1

6. 𝟏+𝒂𝒙 𝒏 =𝟏+𝒏 𝒂𝒙 + 𝒏 𝒏−𝟏 𝟐! 𝒂𝒙 𝟐 + 𝒏 𝒏−𝟏 (𝒏−𝟐) 𝟑! 𝒂𝒙 𝟑 + …
WB11b
b) Hence, find the expansion of 2𝑥+7 (𝑥+3)(𝑥+4) up to and including the term in 𝑥 3
2𝑥+7 (𝑥+3)(𝑥+4) = 4+𝑥 − 𝑥 −1
4+𝑥 −1 = 𝑥 −1 =
1 4 1− 1 4 𝑥 𝑥 2 − 𝑥 3 +…
3+𝑥 −1 = 𝑥 −1 =
1 3 1− 1 3 𝑥 𝑥 2 − 𝑥 3 +…
2𝑥+7 (𝑥+3)(𝑥+4) = − 1 16 𝑥 𝑥 2 − 𝑥 3 +… + 1− 1 3 𝑥 𝑥 2 − 𝑥 3 +…
Carefully collect the terms
2𝑥+7 (𝑥+3)(𝑥+4) = − 𝑥 𝑥 2 − 𝑥 3 +…

7. f x = 3 𝑥 2 +18 1−2𝑥 2+𝑥 2 = 𝐴 1−2𝑥 + 𝐵 2+𝑥 + 𝐶 2+𝑥 2 find A, B and C
WB12a
f x = 3 𝑥 −2𝑥 2+𝑥 2 = 𝐴 1−2𝑥 + 𝐵 2+𝑥 + 𝐶 2+𝑥 find A, B and C
Hence, find the expansion of 3 𝑥 (1−2𝑥) 2+𝑥 2 up to and including the term in 𝑥 3
3 𝑥 (1−2𝑥) 2+𝑥 2 = 𝐴 1−2𝑥 + 𝐵 2+𝑥 + 𝐶 2+𝑥 2
3 𝑥 2 +18= 𝐴 𝑥+4 + 𝐵(𝑥+3)
𝐴= 𝐵=0 𝐶=6
3 𝑥 (1−2𝑥) 2+𝑥 2 = 3 1−2𝑥 𝑥 2
Check by adding together!
3 𝑥 (1−2𝑥) 2+𝑥 2 =3 1−2𝑥 − 𝑥 −2
3 𝑥 (1−2𝑥) 2+𝑥 2 =3 1−2𝑥 − 𝑥 −2

8. 𝟏+𝒂𝒙 𝒏 =𝟏+𝒏 𝒂𝒙 + 𝒏 𝒏−𝟏 𝟐! 𝒂𝒙 𝟐 + 𝒏 𝒏−𝟏 (𝒏−𝟐) 𝟑! 𝒂𝒙 𝟑 + …
WB13b
b) Hence, find the expansion of 3 𝑥 (1−2𝑥) 2+𝑥 2 up to and including the term in 𝑥 3
3 𝑥 (1−2𝑥) 2+𝑥 2 =3 1−2𝑥 − 𝑥 −2
3 1−2𝑥 −1 =
3 1+2𝑥+4 𝑥 2 +8 𝑥 3 +…
𝑥 −2 =
3 2 1−𝑥 𝑥 2 − 1 2 𝑥 3 +…
3 𝑥 (1−2𝑥) 2+𝑥 2 = 3+6𝑥+12 𝑥 𝑥 3 +… − 3 2 𝑥 𝑥 2 − 3 4 𝑥 3 +…
Carefully collect the terms
3 𝑥 (1−2𝑥) 2+𝑥 2 = 𝑥 𝑥 𝑥 3 +…

9. One thing to improve is –
KUS objectives
BAT complete problems involving partial fractions followed by series expansion
self-assess
One thing learned is –
One thing to improve is –

10. END