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Ex 8.1,4 - Expand (x/3 + 1/x)5 - Chapter 8 Class 11 CBSE

Ex 8.1,4 - Chapter 8 Class 11 Binomial Theorem - Part 2
Ex 8.1,4 - Chapter 8 Class 11 Binomial Theorem - Part 3


Transcript

Ex 8.1, 4 Expand the expression (𝑥/3+1/𝑥)^5 We know that (a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + ….…. + nCn – 1 a1 bn – 1 + nCn bn Hence (a + b)5 = = 5!/0!( 5 − 0)! a5 + 5!/1!( 5 − 1)! a4 b1 + 5!/2!( 5 − 2)! a3 b2 + 5!/3!( 5 − 3)! a2b3 + 5!/4!( 5 − 4)! a b4 + 5!/5!( 5 −5)! b5 = 5!/(0! × 5!) a5 + 5!/(1! × 4!) a4 b + 5!/(2! 3!) a3 b2 + 5!/(3! 2!) a2b3 + 5!/(4! 1!) a b4 + 5!/(5! 0!) b5 = 5!/5! a5 + (5 × 4!)/4! a4 b + (5 × 4 × 3!)/(2! 3!) a3 b2 + (5 × 4 × 3!)/(2 × 1 ×3!) a3b2 + (5 × 4 × 3!)/(2 ×1 ×3!) a2b3 + (5 × 4!)/4! ab4 + 5!/(5! ) b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 We need to find (𝑥/3+1/𝑥)^5 Putting a = 𝑥/3 & b = 1/𝑥 (𝑥/3+1/𝑥)^5 = (x/3)^5+ 5 (x/3)^4 (1/x) + 10 (x/3)^3 (1/x)^2 + 10 (x/3)^2 (1/x)^3 + 5 (x/3) (1/x)^4 + (1/x)^5 = 𝑥5/243 + 5 (𝑥^4/81)(1/𝑥) + 10(𝑥^3/27) (1/𝑥)^2 + 10 (𝑥^2/9)(1/𝑥^3 ) + 5 (𝑥/3) (1/𝑥^4 ) +(1/𝑥^5 ) = 𝒙𝟓/𝟐𝟒𝟑 + 𝟓/𝟖𝟏 𝒙3 + 𝟏𝟎/𝟐𝟕 𝒙 + 𝟏𝟎/𝟗𝒙 + 𝟓/𝟑𝒙𝟑 + 𝟏/𝒙𝟓

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.