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  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise

Transcript

Ex 8.1, 4 Expand the expression (๐‘ฅ/3+1/๐‘ฅ)^5 We know that (a + b)n = nC0 an + nC1 an โ€“ 1 b1 + nC2 an โ€“ 2 b2 + โ€ฆ.โ€ฆ. + nCn โ€“ 1 a1 bn โ€“ 1 + nCn bn Hence (a + b)5 = = 5!/0!( 5 โˆ’ 0)! a5 + 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 = 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!)/(2 ร—1 ร—3!) a2b3 + (5 ร— 4!)/4! ab4 + 5!/(5! ) b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 We need to find (๐‘ฅ/3+1/๐‘ฅ)^5 Putting a = ๐‘ฅ/3 & b = 1/๐‘ฅ (๐‘ฅ/3+1/๐‘ฅ)^5 = (x/3)^5+ 5 (x/3)^4 (1/x) + 10 (x/3)^3 (1/x)^2 + 10 (x/3)^2 (1/x)^3 + 5 (x/3) (1/x)^4 + (1/x)^5 = ๐‘ฅ5/243 + 5 (๐‘ฅ^4/81)(1/๐‘ฅ) + 10(๐‘ฅ^3/27) (1/๐‘ฅ)^2 + 10 (๐‘ฅ^2/9)(1/๐‘ฅ^3 ) + 5 (๐‘ฅ/3) (1/๐‘ฅ^4 ) +(1/๐‘ฅ^5 ) = ๐’™๐Ÿ“/๐Ÿ๐Ÿ’๐Ÿ‘ + ๐Ÿ“/๐Ÿ–๐Ÿ ๐’™3 + ๐Ÿ๐ŸŽ/๐Ÿ๐Ÿ• ๐’™ + ๐Ÿ๐ŸŽ/๐Ÿ—๐’™ + ๐Ÿ“/๐Ÿ‘๐’™๐Ÿ‘ + ๐Ÿ/๐’™๐Ÿ“

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.