Question

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

Answer 1

To Prove : coefficient of xn in (1+x)²ⁿ = 2 × coefficient of xⁿ in (1+x)^2n-1 

For (1+x)²ⁿ , 

a =1, b = x and m = 2n 

We have a formula

To get the coefficient of xn , we must have, 

xⁿ = x^r 

r = n

Therefore, the coefficient of xⁿ = \((\frac{2n}n)\)

Therefore, the coefficient of xⁿ in (1+x)²ⁿ = \(\frac{2\times(2n-1)!}{n!\times(n-1)!}\)......eq(1)

Now for (1+x)^2n-1 , 

a = 1, b = x and m = 2n -1 

We have formula,

To get the coefficient of xn , we must have, 

xⁿ = x^r 

r = n 

Therefore, the coefficient of xⁿ in (1+x)^2n-1 = \((\frac{2n-1}n)\)

…..multiplying and dividing by 2 

Therefore, 

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

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

Hence proved.