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^+?

1 Answer
Mar 31, 2017

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)