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Last updated at Jan. 30, 2020 by Teachoo

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

Ex 8.2

Ex 8.2,1
Not in Syllabus - CBSE Exams 2021
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Ex 8.2, 2 Important Not in Syllabus - CBSE Exams 2021

Ex 8.2, 3 Not in Syllabus - CBSE Exams 2021

Ex 8.2, 4 Important Not in Syllabus - CBSE Exams 2021

Ex 8.2,5 Not in Syllabus - CBSE Exams 2021

Ex 8.2 6 Important Not in Syllabus - CBSE Exams 2021

Ex 8.2,7 Not in Syllabus - CBSE Exams 2021

Ex 8.2,8 Important Not in Syllabus - CBSE Exams 2021

Ex 8.2,9 Not in Syllabus - CBSE Exams 2021

Ex 8.2,10 Important Not in Syllabus - CBSE Exams 2021

Ex 8.2,11 Important Not in Syllabus - CBSE Exams 2021

Ex 8.2,12 Not in Syllabus - CBSE Exams 2021

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.