# Example 16 - Class 11

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 16 The sum of the coefficients of the first three terms in the expansion of (x – 3/x^2)^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–rbr for (x – 3𝑥2)m Putting n = m , a = x , b = −3𝑥2 Tr+1 = mCr xm–r (−3𝑥2)r Now we need to find first 3 terms i.e. T1 , T2 and T3 i.e. T0+1 , T1+1 and T2+1 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 × 𝑚𝑚−1! 𝑚−1! + 9 × 𝑚 𝑚−1 𝑚−2!2 × 𝑚−2! = 559 1 – 3m + 92 m(m–1) = 559 21−2(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 Finding general term Tr+1 = mCr xm – r −3𝑥2𝑟 = 12Cr x12 – r −3𝑥2𝑟 = 12Cr (x)12 – r ( –3)r 1𝑥2𝑟 = 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 Comparing powers 3 = 12 – 3r 3r = 12 – 3 3r = 9 r = 93 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

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.