Example 13 - Chapter 8 Class 11 Binomial Theorem (Deleted)
Last updated at Jan. 29, 2020 by Teachoo
Last updated at Jan. 29, 2020 by Teachoo
Transcript
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 an +nC1 an – 1 b1 + …. …. + nCn – 1 a1 bn – 1+ nCn bn Hence (a + b)4 = 4C0 a4 + 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)!) ab3 + 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, (a + b)n = nC0 an + nC1 an – 1 b1 + ….……. + nCn – 1 a1 bn – 1 + nCn bn (a + b)5 = 5C0 a5+ 5C1 a4 b1 + 5C2 a3 b2 + 5C3 a2b3 + 5C4 a b4 + 5C5 b5 = 5!/0!(5 − 0)! a5 + 5!/1!(5 − 1)! a4 b + 5!/2!(5 − 2)! a3 b2 + 5!/3!(5 − 3)! a2b3 + 5!/4!(5 − 4)! ab4 + 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 × 1 ×3!) a3b2 + (5 × 4 × 3!)/(2 × 1 ×3!) a2b3 + (5 × 4!)/4! ab4 + 5!/(5! ) b5 = a5 + 5a4 b + 10 a3 b2 + 10 a2b3 + 5ab4 + 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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