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Find the sum of $\binom n0 - \binom n2 +\binom n4 -\binom n6 \cdots$


Using Binomial expansion of $(1+x)^n$, $$ \binom n0 +\binom n1 x^1 + \binom n 2 x^2 + \cdots + \binom n nx^n $$ Substituting $x = i$, $$(1+i)^n = \binom n0 +\binom n1 i^1 + \binom n 2 i^2 + \cdots + \binom n ni^n $$ $$(1+i)^n = \binom n0 +\binom n1 i - \binom n 2 - \binom n 3 i\cdots + \binom n ni^n $$ How to proceed further?

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So you have $$(1+i)^{n}= \binom{n}{0} + \binom{n}{1} i + \binom{n}{2}i^2 + \cdots = \binom{n}{0}-\binom{n}{2} + \binom{n}{4} + i \cdot\left[\binom{n}{1}-\binom{n}{3} + \cdots\right]$$

  • Write $(1+i)^{n}= (\sqrt{2})^{n}\cdot \left(\frac{1}{\sqrt{2}}+i \cdot\frac{1}{\sqrt{2}}\right)^{n}$ and see if you can find the real part of this term.
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    $\begingroup$ How did you get the polar form of $(1+i)^n$ ? $\endgroup$ – Ardent Oct 8 '19 at 7:42
  • $\begingroup$ Note that $1+i = \sqrt{2}e^{i \pi/4}$. Hence the polar form of $(1+i)^n$ is $\left(\sqrt{2}\right)^n e ^{i n\pi/4}$. $\endgroup$ – Minus One-Twelfth Oct 8 '19 at 7:54
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Hint: Try taking the real part of $(1+i)^n$ in two different ways.

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Note that $$ \frac12\left(i^k+(-i)^k\right)=\left\{\begin{array}{rl} 1&\text{if }k\equiv0&\pmod4\\ 0&\text{if }k\equiv1&\pmod4\\ -1&\text{if }k\equiv2&\pmod4\\ 0&\text{if }k\equiv3&\pmod4 \end{array}\right. $$ Thus, $$ \begin{align} \sum_{k=0}^n\binom{n}{k}\frac12\left(i^k+(-i)^k\right) &=\frac12\left((1+i)^n+(1-i)^n\right)\\ &=2^{n/2}\frac{e^{in\pi/4}+e^{-in\pi/4}}2\\[9pt] &=2^{n/2}\cos(n\pi/4) \end{align} $$

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