Hi. I need help in the following problem please.
If $\displaystyle (1+z)^n=a_0 + \sum{a_k z^k}$, where $\displaystyle n$ is a positive integer, prove that:
$\displaystyle a_0+a_4+a_8+...=2^{n-2} + 2^{\frac{1}{2}n-1}cos\frac{n\pi}{4}$
Thank you
Hi. I need help in the following problem please.
If $\displaystyle (1+z)^n=a_0 + \sum{a_k z^k}$, where $\displaystyle n$ is a positive integer, prove that:
$\displaystyle a_0+a_4+a_8+...=2^{n-2} + 2^{\frac{1}{2}n-1}cos\frac{n\pi}{4}$
Thank you
I haven't worked out the whole thing but I think that the fact that the indices in the result are multiples of 4 would lead to looking at z= i. Recall that if z= i, then $\displaystyle z^2= -1$, $\displaystyle z^3= -i$, and [itex]z^4= 1[/tex]. That is, taking z= i gives, as real part, $\displaystyle (z_0+ z_4+ z_8+ \cdot\dot\dot)- (z_2+ z_6+ z_{10}+ \cdot\cdot\cdot)$.