Find a positive value of m for which the coefficient of x² in the expansion (1 + x)ᵐ is 6

Solution:

It is known that (r + 1)ᵗʰ term, (Tᵣ ₊ ₁) in the binomial expression of (a + b)ⁿ is given by Tᵣ ₊ ₁ = ⁿCᵣ aⁿ ⁻ ʳ bʳ

Assuming that xᵐ occurs in the (r + 1)ᵗʰ term of the expansion (1 + x)ᵐ, we obtain

Tᵣ ₊ ₁ = ᵐCᵣ (1)ᵐ ⁻ ʳ (x)ʳ

= ᵐCᵣ xʳ

Comparing the indices of x in x and in x² is ᵐC₂

ᵐC₂ = 6

m!/2!(m - 2)! = 6

[(m (m - 1) x (m - 2)!]/[2 x (m - 2)!] = 6

m (m - 1) = 12

m² - m - 12 = 0

m² - 4m + 3m - 12 = 0

m (m - 4) + 3(m - 4) = 0

(m - 4)(m + 3) = 0

m = 4 or m = - 3

Thus, 4 is the positive value of m for which the coefficient of x² in the expansion (1 + x)ᵐ is 6

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NCERT Solutions Class 11 Maths Chapter 8 Exercise 8.2 Question 12

Find a positive value of m for which the coefficient of x² in the expansion (1 + x)ᵐ is 6

Summary:

Using binomial theorem, we have to find a positive value of m for which the coefficient of x² in the expansion (1 + x)ᵐ is 6. We have found that 4 is the positive value of m for which the coefficient of x² in the expansion (1 + x)ᵐ is 6