Example 16 - The sum of coefficients of first 3 terms in (x - 3/x^2)^m

Example 16 - Chapter 8 Class 11 Binomial Theorem - Part 2
Example 16 - Chapter 8 Class 11 Binomial Theorem - Part 3
Example 16 - Chapter 8 Class 11 Binomial Theorem - Part 4
Example 16 - Chapter 8 Class 11 Binomial Theorem - Part 5
Example 16 - Chapter 8 Class 11 Binomial Theorem - Part 6
Example 16 - Chapter 8 Class 11 Binomial Theorem - Part 7
Example 16 - Chapter 8 Class 11 Binomial Theorem - Part 8

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

Transcript

Example 16 The sum of the coefficients of the first three terms in the expansion of (x – 3/x2)m , x β‰  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 – r br For (x – πŸ‘/π’™πŸ)m Putting n = m , a = x , b = (βˆ’3)/π‘₯^2 Tr+1 = mCr xm – r ((βˆ’3)/π‘₯^2 )^π‘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 m–0 ((βˆ’3)/π‘₯2)^0 = mC0 xm Γ— 1 = mC0 xm = 1 (xm) ∴ Coefficient of first term = 1 For T2 T1+1 = mC1 x m–1 ((βˆ’3)/π‘₯2)^1 = mC1 (x) m–1 Γ— (-3) Γ— 1/π‘₯2 = mC1 (x) m–1 Γ— (-3) Γ— xβˆ’2 = –3 mC1 . Xm – 1 – 2 = –3 mC1 .xm – 3 ∴ Coefficient of second term = –3 mC1 For T3 T2+1 = mC2 (x) m–2 ((βˆ’3)/π‘₯2)^2 = mC2 (x) m – 2 (βˆ’3)2 (1/π‘₯2)^2 = mC2 (x) m – 2 Γ— (9) Γ— xβˆ’4 = mC2 . 9 .xm – 2 – 4 = 9 mC2 .xm – 6 Coefficient of third term = 9 mC2 Given that Sum of coefficients of first three terms = 559 1 + ( –3 mC1 ) + 9 mC2 = 559 1 – 3 Γ— π‘š!/1!(π‘š βˆ’ 1)! + 9 Γ— π‘š!/2!(π‘š βˆ’ 2)! = 559 1 – 3 Γ— π‘š(π‘š βˆ’ 1)!/( (π‘š βˆ’ 1)!) + 9 Γ— (π‘š (π‘š βˆ’ 1)(π‘š βˆ’ 2)!)/(2 Γ— (π‘š βˆ’ 2)!) = 559 1 – 3m + 9/2 m(m – 1) = 559 (2 βˆ’ 2(3π‘š) + 9π‘š (π‘š βˆ’ 1))/2 = 559 (2 βˆ’ 6π‘š + 9π‘š (π‘š βˆ’ 1))/2 = 559 2 – 6m + 9m2 – 9m = 559 Γ— 2 2 – 15m + 9m2 = 1118 9m2 – 15m + 2 – 1118 = 0 9m2 – 15m – 1116 = 0 3(3m2 – 5m – 372) = 0 3m2 – 5m – 372 = 0 3m2 – 36m + 31m – 372 = 0 3m (m – 12) + 31 (m – 12) = 0 (3m + 31) (m – 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 = –31 m = (βˆ’31)/3 m – 12 = 0 m = 12 Finding general term Tr+1 = mCr xm – r ((βˆ’3)/π‘₯2)^π‘Ÿ = 12Cr x12 – r ((βˆ’3)/π‘₯2)^π‘Ÿ = 12Cr (x)12 – r ( –3)r (1/π‘₯2)^π‘Ÿ = 12Cr (x)12 – r ( –3)r x-2r = 12Cr . (–3)r .x12– r – 2r = 12Cr . (–3)r .x12 – 3r We need the term containing x3 ∴ x3 = x12 – 3r (Putting m = 12) Comparing powers 3 = 12 – 3r 3r = 12 – 3 3r = 9 r = 9/3 r = 3 Hence, the term containing x3 Tr +1 = 12Cr . (–3)r . x12 – 3r T3+1 = 12C3 (–3)3 . x12 – 3(3) = 12!/3!(12 βˆ’13)! (–3)3 . x3 = 12!/(3! 9!) (–27).x3 = (12 Γ— 11 Γ— 10 Γ— 9!)/(3 Γ— 2 Γ— 1 Γ— 9!) (–27) x3 = (12 Γ— 11 Γ— 10 )/(3 Γ— 2 ) (–27) x3 = (12 Γ— 11 Γ— 10 )/(3 Γ— 2 ) (–27) x3 = – 5940 x3 Hence the Required term T4 is –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 10 years. He provides courses for Maths and Science at Teachoo.