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 .
It
is known that (r
+ 1)th
term, (Tr+1),
in the binomial expansion of (a +
b)n
is given by
.
Assuming that xn occurs in the (r + 1)th term of the expansion of (1 + x)2n, we obtain
Comparing the indices of x in xn and in Tr + 1, we obtain
r = n
Therefore, the coefficient of xn in the expansion of (1 + x)2n is
Assuming that xn occurs in the (k +1)th term of the expansion (1 + x)2n – 1, we obtain
Comparing the indices of x in xn and Tk + 1, we obtain
k = n
Therefore, the coefficient of xn in the expansion of (1 + x)2n –1 is
From (1) and (2), it is observed that
Therefore, the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1.
Hence, proved.