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Ex 7.1, 1 Expand the expression (1 – 2x)5 (1 – 2x)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!)/(3! ×1 ×3!) a2b3 + (5 × 4!)/4! ab4 + 5!/(5! ) b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 Thus, (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 Putting a = 1 & b = (– 2x) (1 – 2x)5 = (1)5 + 5(1)4 (–2x) + 10 (1)3 (–2x)3 + 10 (1)2 (–2x)3 + 5 (1) (–2x)4 + (–2x)5 = 1 – 10x + 10(4x2) + 10 (–8x3) + 5 (16x4) + (–32x5) = 1 – 10x + 40x2 – 80x3 + 80x4 – 32x5

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.