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I am trying to show that if $$\left( 1+x\right) ^{n}=p_{0}+p_{1}x+p_{2}x^{2}+\ldots $$ and n being a positive integer, then $$p_{0}-p_{2}+p_{4}+\ldots = 2^{\frac {n} {2} }\cos \dfrac {n\pi } {4}$$ and $$p_{1}-p_{3}+p_{5}+\ldots = 2^{\frac {n} {2} }\sin \dfrac {n\pi } {4}$$.

Solution Attempt

First i thought of using $$ (1 + x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r$$ as in getting the values of the coefficient so $$p_{0}-p_{2}+p_{4}+\ldots = 1-\dfrac {n!} {2!\left( n-2\right) !}+\dfrac {n!} {4!\left( n-4\right) !}-\dfrac {n!} {6!\left( n-6\right) !}$$

I do n't know how to deal with the n! and n related factorial terms in the denominator,The two algebraic manipulations which come to mind both seem fruitless.

Any help with a solution or proof strategy would be much appreciated.

PS : If i was to assume all the p coefficients are complex numbers and wanted to write them using the cos and sin version is there any mathematical relationship between consecutive p's theta values ?

so if $p_{0} =\cos\theta_{0} + i\sin\theta_{0}$ and $p_{1} =\cos\theta_{1} + i\sin\theta_{1}$ is there a relationship between $\theta_{0}$ and $\theta_{1}$ ?

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up vote 2 down vote accepted


See about the multisection of a serie. If you use modulus 4, probably you will can find the result.

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Hint: Let $x=it$. Expand. Then let $x=-it$. Expand. Add and subtract.

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Succinctness FTW. – Pedro Tamaroff Apr 23 '12 at 1:18

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