# Misc 5 - Chapter 8 Class 11 Binomial Theorem (Deleted)

Last updated at Jan. 29, 2020 by Teachoo

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

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 You are here

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

Last updated at Jan. 29, 2020 by Teachoo

Misc 5 Evaluate (√3 + √2 )6 – (√3 – √2 )6 . Finding (a + b)6 – (a – b)6 We know that (a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + ….…. + nCn – 1 a1 bn – 1 + nCn bn Hence (a + b)6 = = 6!/(0! (6 − 0)!) a6 × 1 + 6!/(1! (6 − 1) !) a5 b + 6!/2!(6 − 2)! a4 b2 + 6!/3!(6 − 3)! a3 b3 + 6!/(4! (6 − 4) !)a2 b4 + 6!/5!(6 − 5)! ab5 + 6!/(6 ! (6 − 6) !) b6 = 6!/(1 × 6! ) a6 + 6!/(1 × 5!) a5 b + 6!/(2! × 4!) a4 b2 + 6!/(3! 3!) a3 b3 + 6!/(4! 2!) a2 b4 + 6!/(5! × 1) a b5 + 6!/(6! × 1) b6 = 6!/6! a6 + (6 ×5!)/(5! ) a5b + (6 × 5 × 4!)/(2 × 4!) a4 b2 + (6 × 5 × 4 × 3!)/(3 × 2 × 1 × 3!) a3 b3 + (6 × 5 × 4!)/(2 × 1 × 4!) a2 b4 + (6 × 5!)/(1 × 5!) ab5 + 6!/6! b6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 For (a – b)6 Replace b by (–b) in (1) (a + (–b)) 6 = a6 + 6a5 (–b) + 15a4 (– b)2 + 20a3 (– b)3 + 15a2 (– b)4 + 6a(– b)5 + (– b)6 (a – b)6 = a6 – 6a5 b + 15a4 b2 – 20a3 b3 + 15a2 b4 – 6ab5 + b6 Now, (a + b)6 − (a – b)6 = (a6 + 6a5 b + 15a4 b2 + 20a3 b3 + 15a2 b4 + 6ab5 + b6) − (a6 – 6a5 b + 15a4 b2 – 20a3 b3 + 15a2 b4 – 6ab5 + b6) = a6 − a6 + 15a4 b2 − 15a4 b2 + 15a2 b4 − 15a2 b4 + b6 − b6 + 6a5 b + 6a5b + 20a3 b3 + 20a3 b3 + 6ab5 + 6ab5 = 12a5 b + 40a3 b3 + 12ab5 = 4ab (3a4 + 10a2b2 + 3b4) Thus, (a + b)6 – (a – b)6 = 4ab (3a4 + 10a2b2 + 3b4) We need to find (√3 + √2)6 – (√3 – √2)6 Putting a = √3 and b =√2 (√3 + √2)6 – (√3 – √2)6 = 4(√3 )(√2 ) (3(√3 )^4+10(√3 )^2 (√2 )^2+3(√2 )^4 ) = 4(√(3×2) ) (3(√3 )^4+10(√3 )^2 (√2 )^2+3(√2 )^4 ) = 4(√6 ) (3(√3 )^4+10(√3 )^2 (√2 )^2+3(√2 )^4 ) = 4(√6 ) (3(3)^(1/2 × 4)+10(3)^(1/2 × 2) (2)^(1/2 × 2)+3(2)^(1/2 × 4) ) = 4(√6 ) (3(3)^2+10(3)^1 (2)^1+3(2)^2 ) = 4(√6 ) (27+60+12) = 4(√6 ) (99) = 396√6 Thus, (√3 + √2)6 – (√3 – √2)6 = 3𝟗𝟔√𝟔