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Transcript

Question 12 Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6. We know that General term of expansion (a + b)n is Tr+1 = nCr an–r br General term of (1 + x)m is Putting n = m, a = 1, b = x Tr + 1 = mCr (1)m-r . (x)r = mCr (1) . (x)r = mCr xr We need coefficient of x2 So putting r = 2 Tr + 2 = mC2 x2 Coefficient of x2 is mC2 It is given that coefficient of x2 is 6 i.e. mC2 = 6 𝑚!/2!(𝑚 − 2)! = 6 (𝑚(𝑚 − 1)(𝑚 − 2)!)/2!(𝑚 − 2)! = 6 (𝑚 (𝑚 − 1))/2 = 6 m (m – 1) = 12 m2 – m = 12 m2 – m – 12 = 0 m2 – 4m + 3m – 12 = 0 m(m – 4) + 3 (m – 4) = 0 (m + 3) (m – 4) = 0 So, m = –3 or m = 4 But we need positive value of m Hence, m = 4

  1. Chapter 7 Class 11 Binomial Theorem
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo