IIT 79 Binomial Series

How did you get that?

(1-x)²ⁿ = C₀ - C₁x + C₂ x² .....

differentiate w.r.t x

2n(1-x)^2n-1 = C₁ - 2C₂x + ...... "1"

again,

(1+x)²ⁿ =

C₀ x²ⁿ + C₁ x^2n-1 ............... "2"

multiplying 1 and 2 we get the required series which in turn is the co efficient on x^2n-1

in the expansion of 2n(1-x)^2n-1 (1+x)²ⁿ

;)

I tried but did not get how to find the coefficient of x^2n-1 in that expansion. Can you help?

co eff of x^2n-1 in 2n(1-x)^2n-1(1+x)^2n-1 (1+x )

= co eff of x^2n-1 in 2n(1-x²)^2n-1 (1+x )

= co eff of x^2n-1 in 2n(1-x²)^2n-1 + 2nx (1- x²)^2n-1

= co eff of x^2n-1 = 2nx(1- x²)^2n-1

= co eff of x^2n-2 in 2n(1-x²)^2n-1