Ex 8.2, 1 - Find coefficient of x5 in (x + 3)8 - Binomial

Ex 8.2,1 - Chapter 8 Class 11 Binomial Theorem - Part 2
Ex 8.2,1 - Chapter 8 Class 11 Binomial Theorem - Part 3

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Question 1 - Introduction Find the coefficient of x5 in (x + 3)8 In 18x5 Coefficient of x5 = 18 Question 1 Find the coefficient of x5 in (x + 3)8 We know that General term of expansion (a + b)n is Tr+1 = nCr an–r br For general term of expansion (x + 3)8 Putting a = x , b = 3 , n = 8 Tr + 1 = 8 Cr x8 – r 3r Now we need to find coefficient of x5 So x8 – r = x5 Comparing powers 8 – r = 5 8 – 5 = r 3 = r r = 3 Putting r = 3 in (1) T3 + 1 = 8 C3 x8 – 3 33 = 8!/(3! 5!) x5 × (3 × 3 × 3) = (8 × 7 × 6 × 5!)/(3! × 5!) × x5 × (3 × 3 × 3) = 1512 x5 Hence coefficient of x5 = 1512

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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