Examples

Chapter 8 Class 11 Binomial Theorem
Serial order wise

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Example 16 The sum of the coefficients of the first three terms in the expansion of (x β 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 β r br For (x β π/ππ)m Putting n = m , a = x , b = (β3)/π₯^2 Tr+1 = mCr xm β r ((β3)/π₯^2 )^π 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 mβ0 ((β3)/π₯2)^0 = mC0 xm Γ 1 = mC0 xm = 1 (xm) β΄ Coefficient of first term = 1 For T2 T1+1 = mC1 x mβ1 ((β3)/π₯2)^1 = mC1 (x) mβ1 Γ (-3) Γ 1/π₯2 = mC1 (x) mβ1 Γ (-3) Γ xβ2 = β3 mC1 . Xm β 1 β 2 = β3 mC1 .xm β 3 β΄ Coefficient of second term = β3 mC1 For T3 T2+1 = mC2 (x) mβ2 ((β3)/π₯2)^2 = mC2 (x) m β 2 (β3)2 (1/π₯2)^2 = mC2 (x) m β 2 Γ (9) Γ xβ4 = mC2 . 9 .xm β 2 β 4 = 9 mC2 .xm β 6 Coefficient of third term = 9 mC2 Given that Sum of coefficients of first three terms = 559 1 + ( β3 mC1 ) + 9 mC2 = 559 1 β 3 Γ π!/1!(π β 1)! + 9 Γ π!/2!(π β 2)! = 559 1 β 3 Γ π(π β 1)!/( (π β 1)!) + 9 Γ (π (π β 1)(π β 2)!)/(2 Γ (π β 2)!) = 559 1 β 3m + 9/2 m(m β 1) = 559 (2 β 2(3π) + 9π (π β 1))/2 = 559 (2 β 6π + 9π (π β 1))/2 = 559 2 β 6m + 9m2 β 9m = 559 Γ 2 2 β 15m + 9m2 = 1118 9m2 β 15m + 2 β 1118 = 0 9m2 β 15m β 1116 = 0 3(3m2 β 5m β 372) = 0 3m2 β 5m β 372 = 0 3m2 β 36m + 31m β 372 = 0 3m (m β 12) + 31 (m β 12) = 0 (3m + 31) (m β 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 = β31 m = (β31)/3 m β 12 = 0 m = 12 Finding general term Tr+1 = mCr xm β r ((β3)/π₯2)^π = 12Cr x12 β r ((β3)/π₯2)^π = 12Cr (x)12 β r ( β3)r (1/π₯2)^π = 12Cr (x)12 β r ( β3)r x-2r = 12Cr . (β3)r .x12β r β 2r = 12Cr . (β3)r .x12 β 3r We need the term containing x3 β΄ x3 = x12 β 3r (Putting m = 12) Comparing powers 3 = 12 β 3r 3r = 12 β 3 3r = 9 r = 9/3 r = 3 Hence, the term containing x3 Tr +1 = 12Cr . (β3)r . x12 β 3r T3+1 = 12C3 (β3)3 . x12 β 3(3) = 12!/3!(12 β13)! (β3)3 . x3 = 12!/(3! 9!) (β27).x3 = (12 Γ 11 Γ 10 Γ 9!)/(3 Γ 2 Γ 1 Γ 9!) (β27) x3 = (12 Γ 11 Γ 10 )/(3 Γ 2 ) (β27) x3 = (12 Γ 11 Γ 10 )/(3 Γ 2 ) (β27) x3 = β 5940 x3 Hence the Required term T4 is β5940 x3