# On the parity of the coefficients of $(x+y)^n$.

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.

• @MohammadZuhairKhan No, my question is different. The key point is the parity. – Wei-Cheng Liu Jan 29 at 9:28
• 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? – Mohammad Zuhair Khan Jan 29 at 14:00
• @MohammadZuhairKhan That is alright. Yes, I will accept the answer below. – Wei-Cheng Liu Jan 30 at 3:08

$$n \choose k$$ 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).