Question

Find a positive value of m for which the coefficient of $x^{2}$ in the expansion $(1 + x)^{m}$ is 6.

Answer 1

Answer:

m = 4

Step-by-step explanation:

(1 + x)^m

Coefficient of x² = mC₂

= m!/(2!(m-2)1)

= m (m - 1) /2

Coefficient of x² =  6

=> m (m - 1) /2 = 6

=> m² - m = 12

=> m² - m - 12 = 0

=> m² - 4m + 3m - 12 = 0

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

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

=> m = 4   , - 3

positive value of m = 4

Answer 2

Answer:

m = 4

Step-by-step explanation:

In this question,

We have been given that  

$(1 + x)^{m}$

Coefficient of x² = $^mC_2$

= $\frac{m!}{(2!(m-2)1)}$

= $\frac{m (m - 1)}{2}$

Coefficient of x² =  6

=> $\frac{m (m - 1)}{2}\ =\ 6$

=> m² - m = 12

=> m² - m - 12 = 0

=> m² - 4m + 3m - 12 = 0

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

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

=> m = 4   , - 3

Positive value of m = 4

Therefore value of m is 4