Ex 8.1, 2 - Expand (2/x - x/2)5 - Chapter 8 Binomial Theorem - Expansion

  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise
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Ex 8.1, 2 Expand the expression (2/๐‘ฅโˆ’๐‘ฅ/2)^5 We know that (a + b)n = nC0 an b0 + nC1 an โ€“ 1 b1 + nC2 an โ€“ 2 b2 + โ€ฆ. โ€ฆ. + nCn โ€“ 1 a1 bn โ€“ 1 + nCn a0 bn Hence (a + b)5 = = 5!/0!( 5 โˆ’ 0)! a5 ร— 1 + 5!/1!( 5 โˆ’ 1)! a4 b1 + 5!/2!( 5 โˆ’ 2)! a3 b2 + 5!/3!( 5 โˆ’ 3)! a2b3 + 5!/4!( 5 โˆ’ 4)! a b4 + 5!/5!( 5 โˆ’5)! b5 ร— 1 = 5!/(0! ร— 5!) a5 + 5!/(1! ร— 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 ร— 4!)/4! a4 b + (5 ร— 4 ร— 3!)/(2! 3!) a3 b2 + (5 ร— 4 ร— 3!)/(2 ร— 1 ร—3!) a3b2 + (5 ร— 4 ร— 3!)/(3! ร—1 ร—3!) a2b3 + (5 ร— 4!)/4! ab4 + 5!/(5! ) b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 Thus, (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 We need to find (2/๐‘ฅโˆ’๐‘ฅ/2)^5i.e. (2/๐‘ฅ+((โˆ’๐‘ฅ)/2))^5 Putting a = 2/๐‘ฅ & b = (โˆ’๐‘ฅ)/2 (2/๐‘ฅ+((โˆ’๐‘ฅ)/2))^5 = (2/๐‘ฅ)^5 + 5(2/๐‘ฅ)^4 ((โˆ’๐‘ฅ)/2)+ 10 (2/๐‘ฅ)^3 ((โˆ’๐‘ฅ)/2)^2 + 10 (2/๐‘ฅ)^2 ((โˆ’๐‘ฅ)/2)^3 + 5(2/๐‘ฅ) ((โˆ’๐‘ฅ)/2)^4 +((โˆ’๐‘ฅ)/2)^5 = 32/๐‘ฅ5 โ€“ 5 (2/๐‘ฅ)^4 (๐‘ฅ/2) + 10(2/๐‘ฅ)^3 (๐‘ฅ/2)^2โ€“ 10 (2/๐‘ฅ)^2 (๐‘ฅ/2)^3 + 5 (2/๐‘ฅ) (๐‘ฅ/2)^4 + ((โˆ’๐‘ฅ)/2)^5 = 32/x5 โ€“ 5 (2/๐‘ฅ)^3 + 10(2/๐‘ฅ) โ€“ 10 (๐‘ฅ/2) + 5 (๐‘ฅ/2)^3โ€“ ๐‘ฅ5/32 = 32/๐‘ฅ5 โ€“ 40/๐‘ฅ3 + 20/๐‘ฅ โ€“ 5 ๐‘ฅ + 5๐‘ฅ3/8 โ€“ ๐‘ฅ5/32

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