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Ex 8.2
Ex 8.2, 2 Important Deleted for CBSE Board 2022 Exams
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Ex 8.2, 4 Important Deleted for CBSE Board 2022 Exams
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Ex 8.2
Last updated at Jan. 29, 2020 by Teachoo
Ex 8.2, 7 Find the middle terms in the expansions of ("3 β " π₯3/6)^7 Number of terms = n = 7 Since n is odd there will be two middle termx ((π + 1)/2)^π‘β term & ((π + 1)/2+1)^π‘β 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 β r br = ((7 + 1)/2)^π‘β term = 4th term = ((7 + 1)/2)^π‘β term = 4th term For ("3 β " ππ/π)^π Putting a = 3 , b = ((βπ₯3)/6) , n = 7 Tr + 1 = 7Cr (3)7 β r ("β" π₯3/6)^π Tr + 1 = 7Cr (3)7 β r (β1)r (π₯3/6)^π Tr + 1 = 7Cr (3)7 β r (β1)r (π₯3/6)^π Tr + 1 = 7Cr (3)7 β r (β1)r (1/6)^π π₯^3π Finding 4th term Tr + 1 = 7Cr (3)7 β r (β1)r (1/6)^π π₯^3π Finding 4th term i.e. T4 = T3 + 1 , Hence r = 3 T3+1 = 7C3 (3)7 β 3 (β1)3 (1/6)^3 π₯^(3(3)) T4 = 7C3 . 34 . (β1) . 1/63 . (x)9 = β 7!/3!(7 β 3)! . 34. 1/((2 Γ 3)3) . π₯9 = β 7!/(3! 4!) . 34 . 1/(23 . 33) . π₯9 = β ((7 Γ 6 Γ 5 Γ 4!)/(3 Γ 2 Γ 1 Γ 4!)) . 34/33 . π₯9/23 = (βπππ)/π x9 Finding 5th term Tr + 1 = 7Cr (3)7 β r (β1)r (1/6)^π π₯^3π Finding 5th term i.e. T5 = T4 + 1 , Hence r = 4 T4+1 = 7C4 (3)7 - 4 (β1)4 (1/6)^4 π₯^(3(4)) = 7C4 . 33 . 1/64 . (x)12 = 7!/4!(7 β 4)! . 33 . 1/(2 Γ 3)4 . x12 = 7!/(3! 4!) . 33 . 1/(24 . 34) . π₯12 = ((7 Γ 6 Γ 5 Γ 4!)/(3 Γ 2 Γ 1 Γ 4!)) . 34/34 . π₯12/24 = ππ/ππ x12 Hence, the middle terms are (βπππ)/π x9 & ππ/ππ x12