Ex 7.1

Chapter 7 Class 11 Binomial Theorem
Serial order wise

### Transcript

Ex 7.1, 11 Find (a + b)4 – (a – b)4. Hence, evaluate (√3+√2)^4– (√3−√2)^4 . 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)4 = 4C0 a4 + 4C1 a3 b1 + 4C2 a2 b2 + 4C3 a1b3 + 4C4 b4 = 4!/0!(4 − 0)! a4 + 4!/(1 × (4 − 1)!) a3b1 + 4!/2!(4 − 2)! a2b2 + 4!/(3!(4 − 3)!) a3 b1 + 4!/4!(4 −4)! b4 = 4!/(1 × 4!) a4 + 4!/(1 × 3!) a3 b + 4!/(2! × 2!) a2 b2 + 4!/(3! × 1!) ab3 + 4!/(4! 0!) b4 = a4 + 4a3 b + 6a2 b2 + 4 ab3 + b4 Hence, (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4 ab3 + b4 Replacing b with –b (a + (–b))4 = a4 + 4a3 (–b) + 6a2 (–b)2 + 4a (–b)3 + (–b)4 (a – b)4 = a4 – 4a3 b + 6a2b2 – 4ab3 + b4 We need to find (a + b)4 – (a – b)4 (a + b)4 – (a – b)4 = (a4 + 4a3b + 6a2b2 + 4ab3 + b4 ) – (a4 – 4a3b + 6a2b2 – 4ab3 + b4 ) = a4 + 4a3b + 6a2b2 + 4ab3 + b4 – a4 + 4a3b – 6a2b2 + 4ab3 – b4 = a4 – a4 + 6a2b2 – 6a2b2 + b4 – b4 + 4a3b + 4a3b + 4ab3 + 4ab3 = 8a3b + 8ab3 = 8ab (a2 + b2) Thus, (a + b)4 – (a – b)4 = 8ab (a2 + b2) We need to find (√𝟑+√𝟐)^𝟒– (√𝟑−√𝟐)^𝟒 Putting a = √3 and b = √2 (√3+√2)^4– (√3−√2)^4= 8(√3 )(√2 ) ((√3 )^2 "+ " (√2 )^2 ) = 8√3 √2 (3+2) = 8√3 √2 (5) = (8 × 5)√(3 ×2) = 40 √𝟔

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#### 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, Social Science, Physics, Chemistry, Computer Science at Teachoo.