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The coefficients of $(x+y)^3=x^3+3x^2y+3xy^2+y^3$ are $1$, $3$, $3$ and $1$. They are all odd numbers. Which of the following options has coefficients that are also all odd numbers?

$(\text A) \ \ (x+y)^5$
$(\text B) \ \ (x+y)^7$
$(\text C) \ \ (x+y)^9$
$(\text D) \ \ (x+y)^{11}$
$(\text E) \ \ (x+y)^{13}$


My solution is using Pascal's triangle [1] to calculate all the coefficients of $(\text A)$ to $(\text E)$. Finally, I find that only $(\text B)$ is the correct answer. I want to know if there is a faster way to solve this question.

Reference

[1] Wikipedia contributors, "Pascal's triangle," Wikipedia, The Free Encyclopedia, https://en.wikipedia.org/w/index.php?title=Pascal%27s_triangle&oldid=877891681 (accessed January 29, 2019).

Note

My question is different from "How does Combination formula relates in getting the coefficients of a Binomial Expansion?" The key point here is the parity.

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  • $\begingroup$ @MohammadZuhairKhan No, my question is different. The key point is the parity. $\endgroup$ – Wei-Cheng Liu Jan 29 at 9:28
  • $\begingroup$ My apologies. I assumed that it was about faster techniques of finding the coefficients of the expansion. May I suggest accepting the answer below if it has answered your question? $\endgroup$ – Mohammad Zuhair Khan Jan 29 at 14:00
  • $\begingroup$ @MohammadZuhairKhan That is alright. Yes, I will accept the answer below. $\endgroup$ – Wei-Cheng Liu Jan 30 at 3:08
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$n \choose k$ is odd, for every $k=0,1,..,n$ if and only if $n=2^m-1$ (i.e. $n$ has only 1's in its base 2 expansion).

This is proved using Lucas' theorem

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