Question

Prove that the coefficient of x n in the expansion of (1 + x ) 2 n is twice the coefficient of x n in the expansion of (1 + x ) 2 n –1 .

Answer 1

We need to prove,

Coefficient of x n in the expansion of ( 1+x ) 2n = Twice the coefficient of x n in the expansion of ( 1+x ) 2n−1

We know that ( r+1 ) th term in T r+1 in binomial expansion of ( a+b ) n is

T r+1 = C n r a n−r ( b ) r

On comparing ( 1+x ) 2n with ( a+b ) n , we get a=1, b=x and n=2n

T r+1 = C 2n r ( 1 ) 2n−r ( x ) r = C 2n r ( x )

Comparing the indices of x in x n and in T r+1 , get r=n

Thus, the coefficient of x n in the expansion of ( 1+x ) 2n is

C 2n n = ( 2n )! n!( 2n−n )! = ( 2n )! n!n! = ( 2n! ) ( n! ) 2 (1)

Assume that x n occurs in the ( k+1 ) th terms of the expansion ( 1+x ) 2n−1 ,

T k+1 = C 2n−1 k ( 1 ) 2n−1−k ( x ) k = C 2n−1 k ( x ) k

Comparing the indices of x in x n and T r+1 , get k=n .

Therefore, the coefficient of x n in the expansion of ( x+1 ) 2n−1 is

C 2n−1 n = ( 2n−1 )! n!( 2n−1−n )! = ( 2n−1 )! n!( n−1 )! = 2n( 2n−1 )! 2n×n!( n−1 )! = 1 2 [ ( 2n )! ( n! ) 2 ] (2)

From the equation (1) and (2), it is observed that

1 2 ( C 2n n )= C 2n−1 n ( C 2n n )=2( C 2n−1 n )

Hence, coefficient of x n in the expansion of ( x+1 ) 2n =2 and the coefficient of x n in the expansion of ( x+1 ) 2n−1 proved.