Get live Maths 1-on-1 Classs - Class 6 to 12

Examples

Example 1

Example 2 Important

Example 3 Important

Example 4

Example 5 Important Deleted for CBSE Board 2023 Exams

Example 6 Important Deleted for CBSE Board 2023 Exams

Example 7 Deleted for CBSE Board 2023 Exams

Example 8 Important Deleted for CBSE Board 2023 Exams

Example 9 Deleted for CBSE Board 2023 Exams

Example 10 Important Deleted for CBSE Board 2023 Exams

Example 11 Important Deleted for CBSE Board 2023 Exams

Example 12 Deleted for CBSE Board 2023 Exams

Example 13 Important Deleted for CBSE Board 2023 Exams

Example 14 Important Deleted for CBSE Board 2023 Exams

Example 15 Important Deleted for CBSE Board 2023 Exams

Example 16 Deleted for CBSE Board 2023 Exams You are here

Example 17 Important Deleted for CBSE Board 2023 Exams

Last updated at March 16, 2023 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