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  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise

Transcript

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Ÿ‘/๐’™๐Ÿ)m Putting n = m , a = x , b = (โˆ’3)/๐‘ฅ^2 Tr+1 = mCr xm โ€“ r ((โˆ’3)/๐‘ฅ^2 )^๐‘Ÿ 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 mโ€“0 ((โˆ’3)/๐‘ฅ2)^0 = mC0 xm ร— 1 = mC0 xm = 1 (xm) โˆด Coefficient of first term = 1 For T2 T1+1 = mC1 x mโ€“1 ((โˆ’3)/๐‘ฅ2)^1 = mC1 (x) mโ€“1 ร— (-3) ร— 1/๐‘ฅ2 = mC1 (x) mโ€“1 ร— (-3) ร— xโˆ’2 = โ€“3 mC1 . Xm โ€“ 1 โ€“ 2 = โ€“3 mC1 .xm โ€“ 3 โˆด Coefficient of second term = โ€“3 mC1 For T3 T2+1 = mC2 (x) mโ€“2 ((โˆ’3)/๐‘ฅ2)^2 = mC2 (x) m โ€“ 2 (โˆ’3)2 (1/๐‘ฅ2)^2 = mC2 (x) m โ€“ 2 ร— (9) ร— xโˆ’4 = 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 + ( โ€“3 mC1 ) + 9 mC2 = 559 1 โ€“ 3 ร— ๐‘š!/1!(๐‘š โˆ’ 1)! + 9 ร— ๐‘š!/2!(๐‘š โˆ’ 2)! = 559 1 โ€“ 3 ร— ๐‘š(๐‘š โˆ’ 1)!/( (๐‘š โˆ’ 1)!) + 9 ร— (๐‘š (๐‘š โˆ’ 1)(๐‘š โˆ’ 2)!)/(2 ร— (๐‘š โˆ’ 2)!) = 559 1 โ€“ 3m + 9/2 m(m โ€“ 1) = 559 (2 โˆ’ 2(3๐‘š) + 9๐‘š (๐‘š โˆ’ 1))/2 = 559 (2 โˆ’ 6๐‘š + 9๐‘š (๐‘š โˆ’ 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 = โ€“31 m = (โˆ’31)/3 m โ€“ 12 = 0 m = 12 Finding general term Tr+1 = mCr xm โ€“ r ((โˆ’3)/๐‘ฅ2)^๐‘Ÿ = 12Cr x12 โ€“ r ((โˆ’3)/๐‘ฅ2)^๐‘Ÿ = 12Cr (x)12 โ€“ r ( โ€“3)r (1/๐‘ฅ2)^๐‘Ÿ = 12Cr (x)12 โ€“ r ( โ€“3)r x-2r = 12Cr . (โ€“3)r .x12โ€“ r โ€“ 2r = 12Cr . (โ€“3)r .x12 โ€“ 3r We need the term containing x3 โˆด 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 . (โ€“3)r . x12 โ€“ 3r T3+1 = 12C3 (โ€“3)3 . x12 โ€“ 3(3) = 12!/3!(12 โˆ’13)! (โ€“3)3 . x3 = 12!/(3! 9!) (โ€“27).x3 = (12 ร— 11 ร— 10 ร— 9!)/(3 ร— 2 ร— 1 ร— 9!) (โ€“27) x3 = (12 ร— 11 ร— 10 )/(3 ร— 2 ) (โ€“27) x3 = (12 ร— 11 ร— 10 )/(3 ร— 2 ) (โ€“27) x3 = โ€“ 5940 x3 Hence the Required term T4 is โ€“5940 x3

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.