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$\binom {20}0 -\binom {20}2+ \binom{20}4-...-\binom{20}{18}+\binom{20}{20}$

The question specifically gives intervals in which the answer is, but it's probably assumed that you should calculate the whole thing.

Now I can with a bit of mental gymnastics get that the number should be negative, but I don't know how to find the answer without brute forcing it.

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Using the binomial theorem, you want $$\Re ((1+i)^{20}) $$ To get this, you can rewrite it in polar form as $$\Re ((\sqrt 2 e^{i\pi/4})^{20}) = 2^{10} \cos(5\pi) = -2^{10}$$

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  • $\begingroup$ Why is it equal to $(1+i)^{20}$? Also, no need to go through the polar form, $(1+i)$ is very generous to exponentiation after being squared. $\endgroup$ – John Doe May 27 '15 at 22:32
  • $\begingroup$ @JohnDoe It's the real part of $(1+i)^{20}$ $\endgroup$ – ogogmad May 27 '15 at 22:34
  • $\begingroup$ @JohnDoe Good point about $(1+i)^2$ $\endgroup$ – ogogmad May 27 '15 at 22:34
  • $\begingroup$ How do you know it's the real part of $(1+i)^{20}$? $\endgroup$ – John Doe May 27 '15 at 22:36
  • $\begingroup$ @JohnDoe Expand $(1+i)^{20}$ using the binomial theorem and break up the real and imaginary parts. Your sum will be the real part. $\endgroup$ – ogogmad May 27 '15 at 22:38
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Hint: One way is to use complex numbers. Consider $$(1+i)^{20}+(1-i)^{20}.\tag{1}$$ Expand using the binomial theorem.

To calculate (1) another way, note that $1+i=\sqrt{2}e^{i\pi/4}$.

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My suggestion is to look at Pascal's Triangle and play with the first couple of rows and see if you can find the pattern of adding up entries along the rows (of course paying attention to the alternating sign).

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    $\begingroup$ This seems more like a comment than an answer. $\endgroup$ – Hayden May 27 '15 at 22:21
  • $\begingroup$ What's the etiquette here? Is it frowned upon to nudge someone in the right direction? $\endgroup$ – Bo Rel May 28 '15 at 0:17
  • $\begingroup$ No, providing hints is fine. To me your hint isn't quite substantial enough to be considered a proper answer; true, this is subjective, but if one simply looks at the sums you get for the first few rows, I doubt that they'll be able to see what the pattern actually is. $\endgroup$ – Hayden May 28 '15 at 1:26
  • $\begingroup$ Oh wow, I misread the problem. I thought they wanted to sum the entire row with alternating signs, which is of course zero. That's why I thought my nudge was a good one. $\endgroup$ – Bo Rel May 28 '15 at 13:45

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