# Ex 8.1,2 - 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 You are here

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

Last updated at Jan. 29, 2020 by Teachoo

Ex 8.1, 2 Expand the expression (2/π₯βπ₯/2)^5 We know that (a + b)n = nC0 an + nC1 an β 1 b1 + nC2 an β 2 b2 + β¦.β¦. + nCn β 1 a1 bn β 1 + nCn bn Hence (a + b)5 = = 5!/0!( 5 β 0)! a5 + 5!/1!( 5 β 1)! a4 b1 + 5!/2!( 5 β 2)! a3 b2 + 5!/3!( 5 β 3)! a2b3 + 5!/4!( 5 β 4)! a b4 + 5!/5!( 5 β5)! b5 = 5!/(0! Γ 5!) a5 + 5!/(1! Γ 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 Γ 4!)/4! a4 b + (5 Γ 4 Γ 3!)/(2! 3!) a3 b2 + (5 Γ 4 Γ 3!)/(2 Γ 1 Γ3!) a3b2 + (5 Γ 4 Γ 3!)/(3! Γ1 Γ3!) a2b3 + (5 Γ 4!)/4! ab4 + 5!/(5! ) b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 We need to find (2/π₯βπ₯/2)^5i.e. (π/π+((βπ)/π))^π Putting a = 2/π₯ & b = (βπ₯)/2 (2/π₯+((βπ₯)/2))^5 = (2/π₯)^5 + 5(2/π₯)^4 ((βπ₯)/2)+ 10 (2/π₯)^3 ((βπ₯)/2)^2 + 10 (2/π₯)^2 ((βπ₯)/2)^3 + 5(2/π₯) ((βπ₯)/2)^4 +((βπ₯)/2)^5 = 32/π₯5 β 5 (2/π₯)^4 (π₯/2) + 10(2/π₯)^3 (π₯/2)^2β 10 (2/π₯)^2 (π₯/2)^3 + 5 (2/π₯) (π₯/2)^4 + ((βπ₯)/2)^5 = 32/x5 β 5 (2/π₯)^3 + 10(2/π₯) β 10 (π₯/2) + 5 (π₯/2)^3β π₯5/32 = ππ/ππ β ππ/ππ + ππ/π β 5π + πππ/π β ππ/ππ