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Last updated at Jan. 29, 2020 by Teachoo
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Example 16 The sum of the coefficients of the first three terms in the expansion of (x โ 3/x2)m , x โ 0, m being a natural number is 559. Find the term of the expansion containing x3. We know that General term of expansion (a + b)n is Tr + 1 = nCr an โ r br For (x โ ๐/๐๐)m Putting n = m , a = x , b = (โ3)/๐ฅ^2 Tr+1 = mCr xm โ r ((โ3)/๐ฅ^2 )^๐ Now we need to find first 3 terms i.e. T1 , T2 and T3 i.e. T0 + 1 , T1 + 1 and T2 + 1 For T1 T0 +1 = mC0 x mโ0 ((โ3)/๐ฅ2)^0 = mC0 xm ร 1 = mC0 xm = 1 (xm) โด Coefficient of first term = 1 For T2 T1+1 = mC1 x mโ1 ((โ3)/๐ฅ2)^1 = mC1 (x) mโ1 ร (-3) ร 1/๐ฅ2 = mC1 (x) mโ1 ร (-3) ร xโ2 = โ3 mC1 . Xm โ 1 โ 2 = โ3 mC1 .xm โ 3 โด Coefficient of second term = โ3 mC1 For T3 T2+1 = mC2 (x) mโ2 ((โ3)/๐ฅ2)^2 = mC2 (x) m โ 2 (โ3)2 (1/๐ฅ2)^2 = mC2 (x) m โ 2 ร (9) ร xโ4 = mC2 . 9 .xm โ 2 โ 4 = 9 mC2 .xm โ 6 Coefficient of third term = 9 mC2 Given that Sum of coefficients of first three terms = 559 1 + ( โ3 mC1 ) + 9 mC2 = 559 1 โ 3 ร ๐!/1!(๐ โ 1)! + 9 ร ๐!/2!(๐ โ 2)! = 559 1 โ 3 ร ๐(๐ โ 1)!/( (๐ โ 1)!) + 9 ร (๐ (๐ โ 1)(๐ โ 2)!)/(2 ร (๐ โ 2)!) = 559 1 โ 3m + 9/2 m(m โ 1) = 559 (2 โ 2(3๐) + 9๐ (๐ โ 1))/2 = 559 (2 โ 6๐ + 9๐ (๐ โ 1))/2 = 559 2 โ 6m + 9m2 โ 9m = 559 ร 2 2 โ 15m + 9m2 = 1118 9m2 โ 15m + 2 โ 1118 = 0 9m2 โ 15m โ 1116 = 0 3(3m2 โ 5m โ 372) = 0 3m2 โ 5m โ 372 = 0 3m2 โ 36m + 31m โ 372 = 0 3m (m โ 12) + 31 (m โ 12) = 0 (3m + 31) (m โ 12) = 0 Hence, Since m is natural number. Hence, m = 12 Now, we have to find the term of the expansion containing x3 3m + 31 = 0 3m = โ31 m = (โ31)/3 m โ 12 = 0 m = 12 Finding general term Tr+1 = mCr xm โ r ((โ3)/๐ฅ2)^๐ = 12Cr x12 โ r ((โ3)/๐ฅ2)^๐ = 12Cr (x)12 โ r ( โ3)r (1/๐ฅ2)^๐ = 12Cr (x)12 โ r ( โ3)r x-2r = 12Cr . (โ3)r .x12โ r โ 2r = 12Cr . (โ3)r .x12 โ 3r We need the term containing x3 โด x3 = x12 โ 3r (Putting m = 12) Comparing powers 3 = 12 โ 3r 3r = 12 โ 3 3r = 9 r = 9/3 r = 3 Hence, the term containing x3 Tr +1 = 12Cr . (โ3)r . x12 โ 3r T3+1 = 12C3 (โ3)3 . x12 โ 3(3) = 12!/3!(12 โ13)! (โ3)3 . x3 = 12!/(3! 9!) (โ27).x3 = (12 ร 11 ร 10 ร 9!)/(3 ร 2 ร 1 ร 9!) (โ27) x3 = (12 ร 11 ร 10 )/(3 ร 2 ) (โ27) x3 = (12 ร 11 ร 10 )/(3 ร 2 ) (โ27) x3 = โ 5940 x3 Hence the Required term T4 is โ5940 x3
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