# Ex 8.1,11 - Chapter 8 Class 11 Binomial Theorem

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex8.1, 11 Find (a + b)4 – (a – b)4. Hence, evaluate 3+24– 3−24 . We know that (a + b)n = nC0 an b0 + nC1 an – 1 b1 + nC2 an – 2 b2 + …. …. + nCn – 1 a1 bn – 1 + nCn a0 bn Hence, (a + b)4 = 4C0 a4 (b)0 + 4C1 a3 b1 + 4C2 a2 b2 + 4C3 a1b3 + 4C4 a0 b4 = 4!0! 4 − 0! a4 × 1 + 4!1×(4−1)! a2 b2 + 4!2!4 −2! ab3 + 4!4!(4 − 4)! 1 × b4 = 4!1 ×4! a4 + 4!1 ×3! a3 b + 4!2!(4 − 2)! a2 b2 + 4!3!4 −3! 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 = 0 + 0 + 0 + 8a3b + 8ab3 = 8 a3b + 8ab3 = 8ab ( a2 + b2 ) Thus, (a + b)4 – (a – b)4 = 8ab ( a2 + b2 ) We need to find 3+24– 3−24 Putting a = 3 and b = 2 3+24– 3−24= 83 2 3 2+ 2 2 = 832 3+2 = 832 5 = (8 × 5)3 ×2 = 40 6

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.