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

The general term T_r+1 in the binomial expansion is given by T_r+1 = ⁿC_r a^n-r b^r

The general term for binomial (1+x)²ⁿ is

T_r+1 = ²ⁿC_r x^r …………………..1

To find the coefficient of xⁿ

r = n

T_n+1 = ²ⁿC_n xⁿ

The coefficient of xⁿ = ²ⁿC_n

The general term for binomial (1+x)^2n-1 is

T_r+1 = ^2n-1C_r x^r

To find the coefficient of xⁿ

Putting n = r

T_r+1 = ^2n-1C_r xⁿ

The coefficient of xⁿ = ^2n-1C_n

We have to prove

Coefficient of xⁿ in (1+x)²ⁿ = 2 coefficient of xⁿ in (1+x)^2n-1

Consider LHS = ²ⁿC_n

= \( \dfrac{2!}{n!(2n-n)!}\)

= \( \dfrac{2n!}{n!n!}\)

Consider RHS = \(2 \times ^{2n-1}C_n\)

= \(2\times \dfrac{(2n-1)!}{n!(2n-1-n)!}\)

= \( 2\times \dfrac{(2n-1)!}{n!(n-1)!}\)

Now multiplying and dividing by n we get

= \( 2\times \dfrac{(2n-1)!}{n!(n-1)!}\times \dfrac{n}{n}\)

= \( \dfrac{2n(2n-1)!}{n!n(n-1)!}\)

From above equations LHS = RHS

Hence the proof