# Example 16

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.