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

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- Mar 26th 2011, 02:27 AMmohammadfawazFinite sum question
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 - Mar 26th 2011, 03:30 AMHallsofIvy
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, $\displaystyle (z_0+ z_4+ z_8+ \cdot\dot\dot)- (z_2+ z_6+ z_{10}+ \cdot\cdot\cdot)$.

- Mar 26th 2011, 05:04 AMmohammadfawaz
Thanx...

it actually worked..

I took z = i, then -i, then 1, then -1. the complex part vanishes and I got a system of two equations two unknowns : (z0 + z4 + z8 + ...) and (z2 + z6 + z10 + ... )