Further binomial Series expansion

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

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=3 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

4−5𝑥 (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 = 3 1−𝑥+ 𝑥 2 − 𝑥 3 + … 2 2−𝑥 −1 = 1− 1 2 𝑥 −1 = 1+ 1 2 𝑥+ 1 4 𝑥 2 + 1 8 𝑥 3 +… 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 +… 4−5𝑥 (1+𝑥)(2−𝑥) =2− 7 2 𝑥+ 11 4 𝑥 2 − 25 8 𝑥 3 +…

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=1 𝐵=1 2𝑥+7 (𝑥+3)(𝑥+4) = 1 𝑥+3 + 1 𝑥+4 Check by adding together! 2𝑥+7 (𝑥+3)(𝑥+4) = 𝑥+4 −1 + 𝑥+3 −1 = 4+𝑥 −1 + 3+𝑥 −1

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 −1 + 3+𝑥 −1 4+𝑥 −1 = 1 4 1+ 1 4 𝑥 −1 = 1 4 1− 1 4 𝑥+ 1 16 𝑥 2 − 1 64 𝑥 3 +… 3+𝑥 −1 = 1 3 1+ 1 3 𝑥 −1 = 1 3 1− 1 3 𝑥+ 1 9 𝑥 2 − 1 27 𝑥 3 +… 2𝑥+7 (𝑥+3)(𝑥+4) = 1 4 − 1 16 𝑥+ 1 64 𝑥 2 − 1 256 𝑥 3 +… + 1− 1 3 𝑥+ 1 9 𝑥 2 − 1 27 𝑥 3 +… Carefully collect the terms 2𝑥+7 (𝑥+3)(𝑥+4) = 7 12 − 25 144 𝑥+ 91 1728 𝑥 2 − 337 20736 𝑥 3 +…

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

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 2 +18 (1−2𝑥) 2+𝑥 2 up to and including the term in 𝑥 3 3 𝑥 2 +18 (1−2𝑥) 2+𝑥 2 =3 1−2𝑥 −1 + 3 2 1+ 1 2 𝑥 −2 3 1−2𝑥 −1 = 3 1+2𝑥+4 𝑥 2 +8 𝑥 3 +… 3 2 1+ 1 2 𝑥 −2 = 3 2 1−𝑥+ 3 4 𝑥 2 − 1 2 𝑥 3 +… 3 𝑥 2 +18 (1−2𝑥) 2+𝑥 2 = 3+6𝑥+12 𝑥 2 +24 𝑥 3 +… + 3 2 − 3 2 𝑥+ 9 8 𝑥 2 − 3 4 𝑥 3 +… Carefully collect the terms 3 𝑥 2 +18 (1−2𝑥) 2+𝑥 2 = 9 2 + 9 2 𝑥+ 105 8 𝑥 2 + 93 4 𝑥 3 +…

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 –

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