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

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

Here a = 1, b = x and n = m

Putting the value

T_r+1 = ^mC_r 1^m-r x^r

= ^mC_r x^r

We need coefficient of x²

∴ putting r = 2

T₂₊₁ = ^mC₂ x²

The coefficient of x² = ^mC₂

Given that coefficient of x² = ^mC₂ = 6

⇒

⇒

⇒ m(m-1) = 12

⇒ m²- m - 12 =0

⇒ m²- 4m +3m - 12 =0

⇒ m(m-4) + 3(m-4) = 0

⇒ (m+3)(m - 4)= 0

⇒ m = - 3, 4

we need positive value of m so m = 4