

Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex 8.1, 2 Expand the expression (2/๐ฅโ๐ฅ/2)^5 We know that (a + b)n = nC0 an b0 + nC1 an โ 1 b1 + nC2 an โ 2 b2 + โฆ. โฆ. + nCn โ 1 a1 bn โ 1 + nCn a0 bn Hence (a + b)5 = = 5!/0!( 5 โ 0)! a5 ร 1 + 5!/1!( 5 โ 1)! a4 b1 + 5!/2!( 5 โ 2)! a3 b2 + 5!/3!( 5 โ 3)! a2b3 + 5!/4!( 5 โ 4)! a b4 + 5!/5!( 5 โ5)! 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 ร3!) a3b2 + (5 ร 4 ร 3!)/(3! ร1 ร3!) 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 (2/๐ฅโ๐ฅ/2)^5i.e. (2/๐ฅ+((โ๐ฅ)/2))^5 Putting a = 2/๐ฅ & b = (โ๐ฅ)/2 (2/๐ฅ+((โ๐ฅ)/2))^5 = (2/๐ฅ)^5 + 5(2/๐ฅ)^4 ((โ๐ฅ)/2)+ 10 (2/๐ฅ)^3 ((โ๐ฅ)/2)^2 + 10 (2/๐ฅ)^2 ((โ๐ฅ)/2)^3 + 5(2/๐ฅ) ((โ๐ฅ)/2)^4 +((โ๐ฅ)/2)^5 = 32/๐ฅ5 โ 5 (2/๐ฅ)^4 (๐ฅ/2) + 10(2/๐ฅ)^3 (๐ฅ/2)^2โ 10 (2/๐ฅ)^2 (๐ฅ/2)^3 + 5 (2/๐ฅ) (๐ฅ/2)^4 + ((โ๐ฅ)/2)^5 = 32/x5 โ 5 (2/๐ฅ)^3 + 10(2/๐ฅ) โ 10 (๐ฅ/2) + 5 (๐ฅ/2)^3โ ๐ฅ5/32 = 32/๐ฅ5 โ 40/๐ฅ3 + 20/๐ฅ โ 5 ๐ฅ + 5๐ฅ3/8 โ ๐ฅ5/32
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