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Question

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

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Solution

The given expression is ( 1+x ) m .

T r+1 = C n r a nr ( b ) r

Suppose x 2 occurs in ( r+1 ) th term of the expansion ( 1+x ) m ,

T r+1 = C m r ( 1 ) mr ( 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!( m2 )! =6 m( m1 )( m2 )! 2!( m2 )! =6 m( m1 )=12

Further simplify,

m 2 m12=0 m 2 4m+3m12=0 m( m4 )+3( m4 )=0 ( m4 )( 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


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