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^+#?

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Answer 1 · 2017-03-31T10:48:38

See below.

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)#