Check sibling questions

Ex 8.1,4 - Expand (x/3 + 1/x)5 - Chapter 8 Class 11 CBSE

Ex 8.1,4 - Chapter 8 Class 11 Binomial Theorem - Part 2
Ex 8.1,4 - Chapter 8 Class 11 Binomial Theorem - Part 3


Transcript

Ex 8.1, 4 Expand the expression (π‘₯/3+1/π‘₯)^5 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)5 = = 5!/0!( 5 βˆ’ 0)! a5 + 5!/1!( 5 βˆ’ 1)! a4 b1 + 5!/2!( 5 βˆ’ 2)! a3 b2 + 5!/3!( 5 βˆ’ 3)! a2b3 + 5!/4!( 5 βˆ’ 4)! a b4 + 5!/5!( 5 βˆ’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Ÿ“/πŸπŸ’πŸ‘ + πŸ“/πŸ–πŸ 𝒙3 + 𝟏𝟎/πŸπŸ• 𝒙 + 𝟏𝟎/πŸ—π’™ + πŸ“/πŸ‘π’™πŸ‘ + 𝟏/π’™πŸ“

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.