Ex 8.2

Chapter 8 Class 11 Binomial Theorem
Serial order wise

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### Transcript

Ex 8.2, 6 Find the 13th term in the expansion of ("9x – " 1/(3√x))^18 , x ≠ 0. We know that General term of expansion (a + b)n is Tr + 1 = nCr an – r br We need to calculate 13th term (i.e. T13 = T12+1 ) of expansion ("9x – " 1/(3√x))^18 Putting r = 12 , n = 18 , a = 9x & b = 1/(3√x) T12 + 1 = 18C12 (9x)18 – 12 ((−1)/(3√x))^12 = 18!/(12!(18 − 12)) (9x)6 × ((−1)/3)^12 (1/√x)^12 = 18!/(12! × 6!) 96 . x6 . (-1)12 . (1/3) 12 .(1/𝑥)^(1/2 " ×" 12) = (18 × 17 × 16 × 15 × 14 × 13 × 12!)/(12! ×(6 × 5 × 4 × 3 × 2 × 1)) . (32)6 . x6 . 1 . 1/312 . 1/𝑥6 = (18 × 17 × 16 × 15 × 14 × 13 )/((6 × 5 × 4 × 3 × 2 )) . 312 . x6 . 1/312 . 1/𝑥6 = (18 × 17 × 16 × 15 × 14 × 13 )/((6 × 5 × 4 × 3 × 2 )) . 312/312 . 𝑥6/𝑥6 = 18564 × 1 = 18564

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.