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Ex 8.2
Ex 8.2, 2 Important Deleted for CBSE Board 2023 Exams
Ex 8.2, 3 Deleted for CBSE Board 2023 Exams
Ex 8.2, 4 Important Deleted for CBSE Board 2023 Exams
Ex 8.2,5 Deleted for CBSE Board 2023 Exams
Ex 8.2 6 Important Deleted for CBSE Board 2023 Exams
Ex 8.2,7 Deleted for CBSE Board 2023 Exams
Ex 8.2,8 Important Deleted for CBSE Board 2023 Exams
Ex 8.2,9 Deleted for CBSE Board 2023 Exams
Ex 8.2,10 Important Deleted for CBSE Board 2023 Exams
Ex 8.2,11 Important Deleted for CBSE Board 2023 Exams
Ex 8.2,12 Deleted for CBSE Board 2023 Exams
Last updated at March 22, 2023 by Teachoo
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