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

Question

Philoid · Accepted Answer

(1+x)2n = 2nC0+2nC1x1+2nC2x2 +…. + 2nCnxn

(1+x)2n-1 = 2n-1C0+2n-1C1x1 + 2n-1C2x2 +…. + 2n-1Cnxn

Coefficient of xn in (1+x)2n – 1 is (2n - 1Cn)

2nCn =

=

= 2 …….(i)

Now 2n - 1Cn =

=

=  …….(ii)

From (i) and (ii) we have

2nCn = 2. 2n-1Cn

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