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Misc 3 - Find coefficient of x5 in product (1 + 2x)6 (1 - x)7

Misc  3 - Chapter 8 Class 11 Binomial Theorem - Part 2
Misc  3 - Chapter 8 Class 11 Binomial Theorem - Part 3 Misc  3 - Chapter 8 Class 11 Binomial Theorem - Part 4 Misc  3 - Chapter 8 Class 11 Binomial Theorem - Part 5


Transcript

Misc 3 Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem. We know that (a + b)n = nC0 an + nC1 an – 1 b1 + ….……. + nCn – 1 a1 bn – 1 + nCn bn Hence (a + b)6 = 6C0 a6 + 6C1 a5 b1 + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 ab5 + 6C6 b6 = a6 + 6!/(1 ! (6 − 1) !) a5 b + 6!/2!(6 − 2)! a4 b2 + 6!/(3 !(6 − 3)!) a3 b3 + 6!/(4 ! (6 − 4) !)a2 b4 + 6!/(5 !(6 − 5)!) ab5 + b6 = a6 + 6!/(1 × 5!) a5 b + 6!/(2! × 4!) a4 b2 + 6!/(3 ! 3!) a3 b3 + 6!/(4 ! 2!) a2 b4 + 6!/(5 ! × 1) a b5 + b6 = a6 + (6 × 5!)/(5! ) a5b + (6 × 5 × 4!)/(2 × 4!) a4 b2 + (6 × 5 × 4 × 3!)/(3 × 2 × 1 × 3!) a3 b3 + (6 × 5 × 4!)/(2 × 1 × 4!) a2 b4 + (6 × 5!)/(1 × 5!) ab5 + b6 = a6 + 6a5b + 15 a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 Putting a = 1 & b = 2x (1 + 2x)6 = 16 + 615 (2x) + 15 14 (2x)2 + 20 (1)3 (2x)3 + 15 (1)2 (2x) 4 + 6(1)(2x)5 + (2x)6 = 1 + 6 (2x) + 15 (4x2) + 20 (8x3) + 15 (16x4) + 6 (32x5) + 64x6 = 1 + 12x + 60x2 + 160x3 + 2404 + 192x5 + 64x6 Similarly, We know that (a + b)n = nC0 an + nC1 an – 1 b1 + ….……. + nCn – 1 a1 bn – 1 + nCn bn Therefore, (a + b)7 =7C0 a7 +7C1 a6 b1 +7C2 a5 b2 +7C3 a4 b3 +7C4 a3 b4 +7C5 a2 b5 + 7C6 a1 b6 + 7C7 b7 = a7 + 7!/1!( 7 − 1)! a6b1 + 7!/2!( 7 − 2)! a5b2 + 7!/3!( 7 − 3)! a4 b3 + 7!/4!(7 − 4)! a3b4 + 7!/5!(7 −5)! a2 b5 + 7!/6!(7 − 6)! a1 b6 + b7 = a7 + 7!/1(6)! a6 b1 + 7!/2!(5)! a5 b2 + 7!/3!(4)! a4 b3 + 7!/4!(3)! a3 b4 + 7!/5!(2)! a2b5 + 7!/6!(1)! a b6 + b7 = a7 + (7(6)!)/(6)! a6 b1 + (7(6)(5)!)/2(5)! a5 b2 + (7(6)(5)(4)!)/3!(4)! a4 b3 + (7(6)(5)(4)!)/4!(3)! a3 b4 + (7(6)(5)!)/5!(2) a2b5 + (7(6)!)/6! a b6 + b7 = a7 + 7a6 b1 + 21a5 b2 + 35 a4 b3 + 35 a3 b4 + 21a2b5 + 7 ab6 + b7 Putting a = 1 & b = –x (1 – x)7 = (1)7 + 7(1)6 (-x)1 + 21(1)5 (-x)2 + 35 (1)4 (-x)3 + 35 (1)3(-x)4 + 21(1)2(-x)5 + 7 (1)(-x)6 + (-x)7 = 1 + 7(-x) + 21 (x2) + 35 (-x3) + 35 (x4) + 21(-x5) + 7 (x6) + (-x7) = 1 – 7x + 21x2 – 35 x3 + 35 x4 – 21x5 + 7x6 – x7 Now, (1 + 2x)6 (1 – x)7 = (1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x) × (1 – 7x + 21x2 – 35 x3 + 35 x4 – 21x5 + 7x6 – x7) ∴ Coefficient of x5 in the given product is 171

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.