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Ex 8.1
Ex 8.1,2 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,3 Deleted for CBSE Board 2022 Exams
Ex 8.1,4 Important Deleted for CBSE Board 2022 Exams You are here
Ex 8.1, 5 Deleted for CBSE Board 2022 Exams
Ex 8.1 6 Deleted for CBSE Board 2022 Exams
Ex 8.1,7 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,8 Deleted for CBSE Board 2022 Exams
Ex 8.1,9 Deleted for CBSE Board 2022 Exams
Ex 8.1,10 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,11 Deleted for CBSE Board 2022 Exams
Ex 8.1,12 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,13 Important Deleted for CBSE Board 2022 Exams
Ex 8.1,14 Important Deleted for CBSE Board 2022 Exams
Ex 8.1
Last updated at Jan. 29, 2020 by Teachoo
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