Check sibling questions

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

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.