The given expression is ( 1+x ) m .
T r+1 = C n r a n−r ( b ) r
Suppose x 2 occurs in ( r+1 ) th term of the expansion ( 1+x ) m ,
T r+1 = C m r ( 1 ) m−r ( x ) r = C m r ( x ) r
Comparing the indices of x in x 2 and in T r+1 ,
r=2
Thus, the coefficient of x 2 is C m 2 .
C m 2 =6 m! 2!( m−2 )! =6 m( m−1 )( m−2 )! 2!( m−2 )! =6 m( m−1 )=12
Further simplify,
m 2 −m−12=0 m 2 −4m+3m−12=0 m( m−4 )+3( m−4 )=0 ( m−4 )( m+3 )=0
Thus, we get the value of m=4 or m=−3 .
However, it is given that m is positive. Therefore, m = 4