# Ex 8.2,1 - Chapter 8 Class 11 Binomial Theorem

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 8.2, 1 Find the coefficient of x5 in (x + 3)8 Introduction In 18x5 Coefficient of x5 = 18 Ex 8.2, 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 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) = (8 × 7 × 6 ×3 × 3 × 3)/(3 × 2) x5 = 1512 x5 Hence coefficient of x5 = 1512

About the Author

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.