Example 13 - Find coefficient of a4 in product (1 + 2a)4 (2 - a)5 - Coefficient

  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise
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Example 13 Find the coefficient of a4 in the product (1 + 2a)4 (2 – a)5 using binomial theorem. We know that (a + b)n = nC0 anb0 +nC1 an – 1 b1 + …. …. + nCn – 1 a1 bn – 1+ nCn a0 bn Hence (a + b)4 = 4C0 a4 b0 + 4C1 a3 b1 + 4C2 a2 b2 + 4C3 a1b3 + 4C4 a0 b4 = ﷐4!﷮0!﷐ 4 − 0﷯!﷯ a4 + ﷐4!﷮1!(4 − 1)!﷯ a3 b + ﷐4!﷮2!﷐ 4 − 2﷯!﷯ a2 b2 + ﷐4!﷮3!(4 −3)!﷯ a2b2 + ﷐4!﷮4!(4 − 4)!﷯ b4 = ﷐4!﷮1 × 4!﷯ a4 + ﷐4!﷮3!﷯ a3 b + ﷐4!﷮2! × 2!﷯ a2 b2 + ﷐4!﷮3! × 1!﷯ ab3 + ﷐4!﷮1! 0!﷯ b4 = a4 + ﷐4 × 3!﷮3!﷯ a3b + ﷐4 ×3 ×2!﷮2 ×2!﷯ a2 b2 + ﷐4 × 3!﷮3!﷯ ab3 + ﷐4!﷮4! × 1﷯ b4 = a4 + 4a3 b + 6a2 b2 + 4 ab3 + b4 (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 b4 Putting a = 1 & b = 2a (1 + 2a)4 = 14 + 4 (1)3 (2a) + 6(1)2 (2a)2 + 4(1) (2a)3 + (2a)4 = 1 + 8a + 6(4a2) +4 (8a3) + 16 a4 = 1 + 8a + 24a2 + 32a3 + 16a4 Also, We know that (a + b)n = nC0 an b0 + nC1 an – 1 b1 + ….……. + nCn – 1 a1 bn – 1 + nCn a0bn (a + b)5 = 5C0 a5 b0 + 5C1 a4 b1 + 5C2 a3 b2 + 5C3 a2b3 + 5C4 a b4 + 5C5 a0 b5 = ﷐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 × 1 = ﷐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 × 1 ×3!﷯ a3b2 + ﷐5 × 4 × 3!﷮2 ×1 ×3!﷯ a2b3 + ﷐5 × 4!﷮4!﷯ ab4 + ﷐5!﷮5! ﷯ b5 = a5 + 5a4 b1 + 10 a3 b2 + 10 a2b3 + 5a b4 + b5 ∴ (a + b)5 = a5 + 5a4b + 10 a3b2 + 10a2b3 + 5ab4 + b5 Putting a = 2 & b = – a (2 – a)5 = (2)5 + 5(2)4 (–a) + 10 (2)3 (–a)2 + 10 (2)2 (–a)3 + 5(2) (–a)4 + (–a)5 = 32 – 5a (16) + 10a2 (8) – 10 a3 (4) + 10a4 – a5 = 32 – 80a + 80a2 – 40 a3 + 10a4 – a5 Now (1 + 2a)4 (2 – 9)5 = (1 + 8a + 24a2 + 32a3 + 16a4) (32 – 80a + 80a2 – 40a3 + 10a4 – a5) Coefficient of a4 is possible as follow Therefore, coefficient of a4 is –438 Hence, the coefficient of a4 in the given product is – 438

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