Chapter 8 Class 11 Binomial Theorem

Class 11
Important Questions for exams Class 11

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

### Transcript

Ex 7.1, 4 Expand the expression (π₯/3+1/π₯)^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!)/(2 Γ1 Γ3!) a2b3 + (5 Γ 4!)/4! ab4 + 5!/(5! ) b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 We need to find (π₯/3+1/π₯)^5 Putting a = π₯/3 & b = 1/π₯ (π₯/3+1/π₯)^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 = π₯5/243 + 5 (π₯^4/81)(1/π₯) + 10(π₯^3/27) (1/π₯)^2 + 10 (π₯^2/9)(1/π₯^3 ) + 5 (π₯/3) (1/π₯^4 ) +(1/π₯^5 ) = ππ/πππ + π/ππ π3 + ππ/ππ π + ππ/ππ + π/πππ + π/ππ