Question 5 Find the 4th term in the expansion of (x – 2y)12 . We know that General term of expansion (a + b)n is Tr+1 = nCr an–r br We need to find fourth term i.e. T4 = T3 + 1 of expansion (x – 12y)12 Putting r = 3 , a = x , b = –2y , n = 12 T3 + 1 = 12C3 (x)12 – 3 (–2y)3 = (12 !)/(3! (12 − 3)!) x9 (–2)3 (y)3 = 12!/(3! × 9!) x9 (–8) y3 = (12 × 11 × 10 × 9!)/(3 × 2 × 9 !) × (–8) × x9 y3 = (12 × 11 × 10 × (−8))/6 × x9 y3 = 2 × 11 × 10 × (–8) × x9 y3 = –220 × 8 x9 y3 = –1760 x9 y3

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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.

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