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Ex 8.1, 2 - Expand the expression (2/x - x/2)^5 - Teachoo

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


Transcript

Ex 8.1, 2 Expand the expression (2/π‘₯βˆ’π‘₯/2)^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!)/(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‘₯βˆ’π‘₯/2)^5i.e. (𝟐/𝒙+((βˆ’π’™)/𝟐))^πŸ“ Putting a = 2/π‘₯ & b = (βˆ’π‘₯)/2 (2/π‘₯+((βˆ’π‘₯)/2))^5 = (2/π‘₯)^5 + 5(2/π‘₯)^4 ((βˆ’π‘₯)/2)+ 10 (2/π‘₯)^3 ((βˆ’π‘₯)/2)^2 + 10 (2/π‘₯)^2 ((βˆ’π‘₯)/2)^3 + 5(2/π‘₯) ((βˆ’π‘₯)/2)^4 +((βˆ’π‘₯)/2)^5 = 32/π‘₯5 – 5 (2/π‘₯)^4 (π‘₯/2) + 10(2/π‘₯)^3 (π‘₯/2)^2– 10 (2/π‘₯)^2 (π‘₯/2)^3 + 5 (2/π‘₯) (π‘₯/2)^4 + ((βˆ’π‘₯)/2)^5 = 32/x5 – 5 (2/π‘₯)^3 + 10(2/π‘₯) – 10 (π‘₯/2) + 5 (π‘₯/2)^3– π‘₯5/32 = πŸ‘πŸ/π’™πŸ“ – πŸ’πŸŽ/π’™πŸ‘ + 𝟐𝟎/𝒙 – 5𝒙 + πŸ“π’™πŸ‘/πŸ– – π’™πŸ“/πŸ‘πŸ

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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.