Ex 7.1

Chapter 7 Class 11 Binomial Theorem
Serial order wise

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

### Transcript

Ex 7.1, 2 Expand the expression (2/π₯βπ₯/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 β5)! 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 Γ3!) a3b2 + (5 Γ 4 Γ 3!)/(3! Γ1 Γ3!) a2b3 + (5 Γ 4!)/4! ab4 + 5!/(5! ) b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 We need to find (2/π₯βπ₯/2)^5i.e. (π/π+((βπ)/π))^π Putting a = 2/π₯ & b = (βπ₯)/2 (2/π₯+((βπ₯)/2))^5 = (2/π₯)^5 + 5(2/π₯)^4 ((βπ₯)/2)+ 10 (2/π₯)^3 ((βπ₯)/2)^2 + 10 (2/π₯)^2 ((βπ₯)/2)^3 + 5(2/π₯) ((βπ₯)/2)^4 +((βπ₯)/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 + ((βπ₯)/2)^5 = 32/x5 β 5 (2/π₯)^3 + 10(2/π₯) β 10 (π₯/2) + 5 (π₯/2)^3β π₯5/32 = ππ/ππ β ππ/ππ + ππ/π β 5π + πππ/π β ππ/ππ