# Ex 8.2,7

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .