Ex 8.2,7 - Chapter 8 Class 11 Binomial Theorem (Deleted)

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Ex 8.2, 7 Find the middle terms in the expansions of ("3 – " 𝑥3/6)^7 Number of terms = n = 7 Since n is odd there will be two middle termx ((𝑛 + 1)/2)^𝑡ℎ term & ((𝑛 + 1)/2+1)^𝑡ℎ 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 = ((7 + 1)/2)^𝑡ℎ term = 4th term = ((7 + 1)/2)^𝑡ℎ term = 4th term For ("3 – " 𝒙𝟑/𝟔)^𝟕 Putting a = 3 , b = ((−𝑥3)/6) , n = 7 Tr + 1 = 7Cr (3)7 – r ("–" 𝑥3/6)^𝑟 Tr + 1 = 7Cr (3)7 – r (–1)r (𝑥3/6)^𝑟 Tr + 1 = 7Cr (3)7 – r (–1)r (𝑥3/6)^𝑟 Tr + 1 = 7Cr (3)7 – r (–1)r (1/6)^𝑟 𝑥^3𝑟 Finding 4th term Tr + 1 = 7Cr (3)7 – r (–1)r (1/6)^𝑟 𝑥^3𝑟 Finding 4th term i.e. T4 = T3 + 1 , Hence r = 3 T3+1 = 7C3 (3)7 – 3 (–1)3 (1/6)^3 𝑥^(3(3)) T4 = 7C3 . 34 . (−1) . 1/63 . (x)9 = – 7!/3!(7 − 3)! . 34. 1/((2 × 3)3) . 𝑥9 = – 7!/(3! 4!) . 34 . 1/(23 . 33) . 𝑥9 = – ((7 × 6 × 5 × 4!)/(3 × 2 × 1 × 4!)) . 34/33 . 𝑥9/23 = (−𝟏𝟎𝟓)/𝟖 x9 Finding 5th term Tr + 1 = 7Cr (3)7 – r (–1)r (1/6)^𝑟 𝑥^3𝑟 Finding 5th term i.e. T5 = T4 + 1 , Hence r = 4 T4+1 = 7C4 (3)7 - 4 (–1)4 (1/6)^4 𝑥^(3(4)) = 7C4 . 33 . 1/64 . (x)12 = 7!/4!(7 − 4)! . 33 . 1/(2 × 3)4 . x12 = 7!/(3! 4!) . 33 . 1/(24 . 34) . 𝑥12 = ((7 × 6 × 5 × 4!)/(3 × 2 × 1 × 4!)) . 34/34 . 𝑥12/24 = 𝟑𝟓/𝟒𝟖 x12 Hence, the middle terms are (−𝟏𝟎𝟓)/𝟖 x9 & 𝟑𝟓/𝟒𝟖 x12