# Ex 8.2,7 - Chapter 8 Class 11 Binomial Theorem

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 8.2,7 Find the middle terms in the expansions of 3 – 𝑥367 Number of terms = n = 7 Since n is odd there will be two middle term 𝑛 + 12𝑡ℎ term & 𝑛 + 12+1𝑡ℎ term Hence we need to find 4th and 5th term i.e. T4 and T5 We know that general term of (a + b)n is Tr + 1 = nCr an – r br For 3 – 𝑥367 Putting a = 3 , b = −𝑥36 , n = 7 Tr + 1 = 7Cr (3)7–r –𝑥36𝑟 Tr + 1 = 7Cr (3)7 – r (–1)r 𝑥36𝑟 Tr + 1 = 7Cr (3)7 – r (–1)r 𝑥36𝑟 Tr + 1 = 7Cr (3)7 – r (–1)r 16𝑟𝑥3𝑟 Finding 4th term T4 = T3 + 1, Hence r = 3 T3+1 = 7C3 (3)7–3 (–1)3 163𝑥3(3) T4 = 7C3 (3)4. 𝑥3 × –163 = 7C3 (3)4. (x3)3 . −163 = – 7C3 . 34 . 163 . (x)9 = – 7!3!7−3! . 34. 1(2 × 3)3 . 𝑥9 = – 7!3! 4! . 34 .123 . 33 . 𝑥9 = – 7 × 6 × 5 × 4!3 × 2 × 1 × 4! . 3433 . 𝑥923 = – 7 × 6 ×5 3 × 2 . 3 . 𝑥98 = −1058 x9 Similarly, Finding 5th term Tr + 1 = 7Cr (3)7 – r (–1)r 16𝑟𝑥3𝑟 …(1) Finding 5th term i.e. T5 = T4+1 , Hence r = 4 T4+1 = 7C4 (3)7–4 (–1)4 164𝑥3(4) = 7C4 . 34 . 164 . (x)12 = 7!4!7−4! . 34 . 12 × 34 . x12 = 7!3! 4! . 34 .124 . 34 . 𝑥12 = 7 × 6 × 5 × 4!3 × 2 × 1 × 4! . 3434 . 𝑥1224 = 7 × 6 × 5 3 × 2 . 1 . 𝑥1216 = 7 × 6 × 5 × 33 × 2 × 16 . x12 = 3548 x12 Hence the middle terms are −1058 x9 & 3548 x12

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.