# Ex 8.1,4 - Chapter 8 Class 11 Binomial Theorem (Deleted)

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

Expansion

Ex 8.1,1
Deleted for CBSE Board 2022 Exams

Ex 8.1,3 Deleted for CBSE Board 2022 Exams

Ex 8.1, 5 Deleted for CBSE Board 2022 Exams

Ex 8.1,4 Important Deleted for CBSE Board 2022 Exams You are here

Ex 8.1,2 Important Deleted for CBSE Board 2022 Exams

Example 1 Deleted for CBSE Board 2022 Exams

Ex 8.1,11 Deleted for CBSE Board 2022 Exams

Misc 6 Deleted for CBSE Board 2022 Exams

Ex 8.1,12 Important Deleted for CBSE Board 2022 Exams

Misc 5 Important Deleted for CBSE Board 2022 Exams

Ex 8.1,14 Important Deleted for CBSE Board 2022 Exams

Misc 10 Deleted for CBSE Board 2022 Exams

Misc 9 Important Deleted for CBSE Board 2022 Exams

Proving binomial theorem by mathematical induction

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.