Question

By using the binomial expansion of `(1+i)^(2n)` can you prove that: `( ""_0^(2n)) - ( ""_2^(2n)) + ( ""_4^(2n)) - ( ""_6^(2n)) + .... + (-1)^n( ""_(2n)^(2n)) = 2^ncos((npi)/2), n in ZZ^+`?

Answer 1

See below.

Explanation:

According to the complex numbers exponential representation

`x+iy=sqrt(x^2+y^2)e^(i phi)` where `phi=arctan(y/x)`

Here

`1+i=sqrt(2)e^(i pi/4)` and

`(1+i)^(2n) = (sqrt(2)e^(ipi/4))^(2n) = (sqrt(2))^(2n)(e^(ipi/4))^(2n)`

so

`(1+i)^(2n)=2^n e^(i (n pi)/2)`

but according to de Moivre's identity

`e^(i phi) = cos(phi)+isin(phi)` so

`e^(i (n pi)/2)=cos((npi)/2)+i sin((npi)/2)`

but as we know from the series expansion for `(1+x)^(2n)`

`(1+x)^(2n)=sum_(k=0)^(2n)((2n),(k))x^k`

so

`(1+i)^(2n)=sum_(k=0)^(2n)((2n),(k))i^k = sum_(k=0)^(n)(-1)^k((2n),(2k))+i sum_(k=0)^(n)(-1)^k((2n),(2k-1))`

Equating the real components we have

`sum_(k=0)^(n)(-1)^k((2n),(2k)) = 2^ncos((npi)/2)`