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

  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise
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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 b0 + nC1 an – 1 b1 + ….……. + nCn – 1 a1 bn – 1 + nCn a0 bn hence (a + b)6 = 6C0 a6 b0 + 6C1 a5 b1 + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 ab5 + 6C6 a0b6 = ﷐6!﷮0! ﷐6 −0﷯!﷯ a6 × 1 + ﷐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 + ﷐6!﷮6 ! ﷐6 −6﷯ !﷯ 1 × b6 = ﷐6!﷮1 × 6! ﷯ 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 + ﷐6!﷮6 ! × 1﷯ b6 = ﷐6!﷮6!﷯ 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 + ﷐6!﷮6!﷯ b6 = a6 + 6a5b + 15 a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 (a + b)6 = 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 b0 + nC1 an – 1 b1 + ….……. + nCn – 1 a1 bn – 1 + nCn a0 bn Therefore, (a + b)7 =7C0 a7 b0 +7C1 a6 b1 +7C2 a5 b2 +7C3 a4 b3 +7C4 a3 b4 +7C5 a2 b5 + 7C6 a1 b6 + 7C7 a0 b7 = ﷐7!﷮0!﷐7 − 0﷯!﷯ 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 + ﷐7!﷮7!﷐7 − 7﷯!﷯ a0 b7 = ﷐7!﷮1﷐7﷯!﷯ 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 + ﷐7!﷮7!﷐0﷯!﷯ b7 = 1 × 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 + 1 × b7 = a7 + 7a6 b1 + 21a5 b2 + 35 a4 b3 + 35 a3 b4 + 21a2b5 + 7 a b6 + b7 (a + b)7 = a7 + 7a6 b1 + 21a5 b2 + 35 a4 b3 + 35 a3 b4 + 21a2b5 + 7 a b6 + 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 is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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