Question

Question 12 Find a positive value of m for which the coefficient of x^2 in the expansion (1 + x)^m is 6.

Class X1 - Maths -Binomial Theorem Page 171

Answer 1

Use the general term in the expansion of(1+ x)^n is
T_{r+1} = nCr (1)^(n-r)x^r


The general term in the expansion (1+x)^m is
T_{r+1}= mCr.x^r

For coefficient of x^2, put r =2
T_{2+1} = mC2.x^2
Coefficient of x^2 = mC2
Now, given mC2 = 6
m!/2!(m-2)! = 6
m(m-1)(m-2)!/2!(m-2)! = 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 but m ≠ -3
So, m = 4