Ex 8.1,4 - Chapter 8 Class 11 Binomial Theorem (Important Question)
Last updated at Jan. 29, 2020 by Teachoo
Last updated at Jan. 29, 2020 by Teachoo
Transcript
Ex 8.1, 4 Expand the expression (๐ฅ/3+1/๐ฅ)^5 We know that (a + b)n = nC0 an + nC1 an โ 1 b1 + nC2 an โ 2 b2 + โฆ.โฆ. + nCn โ 1 a1 bn โ 1 + nCn bn Hence (a + b)5 = = 5!/0!( 5 โ 0)! a5 + 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 = 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!)/(2 ร1 ร3!) a2b3 + (5 ร 4!)/4! ab4 + 5!/(5! ) b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 We need to find (๐ฅ/3+1/๐ฅ)^5 Putting a = ๐ฅ/3 & b = 1/๐ฅ (๐ฅ/3+1/๐ฅ)^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 = ๐ฅ5/243 + 5 (๐ฅ^4/81)(1/๐ฅ) + 10(๐ฅ^3/27) (1/๐ฅ)^2 + 10 (๐ฅ^2/9)(1/๐ฅ^3 ) + 5 (๐ฅ/3) (1/๐ฅ^4 ) +(1/๐ฅ^5 ) = ๐๐/๐๐๐ + ๐/๐๐ ๐3 + ๐๐/๐๐ ๐ + ๐๐/๐๐ + ๐/๐๐๐ + ๐/๐๐
Chapter 8 Class 11 Binomial Theorem
Ex 8.1,4 Important Deleted for CBSE Board 2022 Exams You are here
Example 5 Important Deleted for CBSE Board 2022 Exams
Ex 8.2,10 Important Deleted for CBSE Board 2022 Exams
Ex 8.2,12 Deleted for CBSE Board 2022 Exams
Example 13 Important Deleted for CBSE Board 2022 Exams
Example 15 Important Deleted for CBSE Board 2022 Exams
Misc 1 Important Deleted for CBSE Board 2022 Exams
Misc 8 Important Deleted for CBSE Board 2022 Exams
Chapter 8 Class 11 Binomial Theorem
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