# Question 12 - Examples - Chapter 7 Class 11 Binomial Theorem

Last updated at April 16, 2024 by Teachoo

Examples

Example 1

Example 2 Important

Example 3 Important

Example 4

Question 1 Important

Question 2 Important

Question 3

Question 4 Important

Question 5

Question 6 Important

Question 7 Important

Question 8

Question 9 Important

Question 10 Important

Question 11 Important

Question 12 You are here

Question 13 Important

Chapter 7 Class 11 Binomial Theorem

Serial order wise

Last updated at April 16, 2024 by Teachoo

Example 16 The sum of the coefficients of the first three terms in the expansion of (x – 3/x2)m , x ≠ 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 – 𝟑/𝒙𝟐)m Putting n = m , a = x , b = (−3)/𝑥^2 Tr+1 = mCr xm – r ((−3)/𝑥^2 )^𝑟 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 ((−3)/𝑥2)^0 = mC0 xm × 1 = mC0 xm = 1 (xm) ∴ Coefficient of first term = 1 For T2 T1+1 = mC1 x m–1 ((−3)/𝑥2)^1 = mC1 (x) m–1 × (-3) × 1/𝑥2 = mC1 (x) m–1 × (-3) × x−2 = –3 mC1 . Xm – 1 – 2 = –3 mC1 .xm – 3 ∴ Coefficient of second term = –3 mC1 For T3 T2+1 = mC2 (x) m–2 ((−3)/𝑥2)^2 = mC2 (x) m – 2 (−3)2 (1/𝑥2)^2 = mC2 (x) m – 2 × (9) × x−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 × 𝑚!/1!(𝑚 − 1)! + 9 × 𝑚!/2!(𝑚 − 2)! = 559 1 – 3 × 𝑚(𝑚 − 1)!/( (𝑚 − 1)!) + 9 × (𝑚 (𝑚 − 1)(𝑚 − 2)!)/(2 × (𝑚 − 2)!) = 559 1 – 3m + 9/2 m(m – 1) = 559 (2 − 2(3𝑚) + 9𝑚 (𝑚 − 1))/2 = 559 (2 − 6𝑚 + 9𝑚 (𝑚 − 1))/2 = 559 2 – 6m + 9m2 – 9m = 559 × 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 = (−31)/3 m – 12 = 0 m = 12 Finding general term Tr+1 = mCr xm – r ((−3)/𝑥2)^𝑟 = 12Cr x12 – r ((−3)/𝑥2)^𝑟 = 12Cr (x)12 – r ( –3)r (1/𝑥2)^𝑟 = 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 −13)! (–3)3 . x3 = 12!/(3! 9!) (–27).x3 = (12 × 11 × 10 × 9!)/(3 × 2 × 1 × 9!) (–27) x3 = (12 × 11 × 10 )/(3 × 2 ) (–27) x3 = (12 × 11 × 10 )/(3 × 2 ) (–27) x3 = – 5940 x3 Hence the Required term T4 is –5940 x3