Slide13.JPG

Slide14.JPG
Slide15.JPG

Go Ad-free

Transcript

Ex 7.1, 5 Expand (x+1/x)^6 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)6 = = 6!/(0! (6 − 0)!) a6 × 1 + 6!/(1! (6 − 1) !) a5 b + 6!/2!(6 − 2)! a4 b2 + 6!/3!(6 − 3)! a3 b3 + 6!/(4! (6 − 4) !)a2 b4 + 6!/5!(6 − 5)! ab5 + 6!/(6 ! (6 − 6) !) b6 = 6!/(1 × 6! ) a6 + 6!/(1 × 5!) a5 b + 6!/(2! × 4!) a4 b2 + 6!/(3! 3!) a3 b3 + 6!/(4! 2!) a2 b4 + 6!/(5! × 1) a b5 + 6!/(6! × 1) b6 = 6!/6! a6 + (6 ×5!)/(5! ) a5b + (6 × 5 × 4!)/(2 × 4!) a4 b2 + (6 × 5 × 4 × 3!)/(3 × 2 × 1 × 3!) a3 b3 + (6 × 5 × 4!)/(2 × 1 × 4!) a2 b4 + (6 × 5!)/(1 × 5!) ab5 + 6!/6! b6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6 We need to find (𝑥 +1/𝑥)^6 Putting a = x & b = 1/𝑥 (𝑥 +1/𝑥)^6 = (x)6 + 6 (x)5 (1/𝑥) + 15 (x4) (1/𝑥)^2 + 20 (x)3 (1/𝑥)^3 + 15 (x)2 (1/𝑥)^4 + 6(x)1 (1/𝑥)^5 + (1/𝑥)^6 = x6 + 6x4 + 15x2 + 20 + 15 × 1/𝑥2 + 6 1/𝑥4 + 1/𝑥6 = x6 + 6x4 + 15x2 + 20 + 𝟏𝟓/𝒙𝟐 + 𝟔/𝒙𝟒 + 𝟏/𝒙𝟔

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo