Mixing
On the parity of the coefficients of $(x+y)^n$.
$begingroup$
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.
binomial-coefficients binomial-theorem
asked Jan 29 at 8:43
Wei-Cheng LiuWei-Cheng Liu
7414
$endgroup$
add a comment |
$begingroup$
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.
binomial-coefficients binomial-theorem
asked Jan 29 at 8:43
Wei-Cheng LiuWei-Cheng Liu
7414
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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.
binomial-coefficients binomial-theorem
asked Jan 29 at 8:43
Wei-Cheng LiuWei-Cheng Liu
7414
$endgroup$
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.
binomial-coefficients binomial-theorem
binomial-coefficients binomial-theorem
asked Jan 29 at 8:43
Wei-Cheng LiuWei-Cheng Liu
7414
asked Jan 29 at 8:43
Wei-Cheng LiuWei-Cheng Liu
7414
edited Jan 29 at 9:31
Wei-Cheng Liu
asked Jan 29 at 8:43
Wei-Cheng LiuWei-Cheng Liu
7414
asked Jan 29 at 8:43
Wei-Cheng LiuWei-Cheng Liu
7414
asked Jan 29 at 8:43
Wei-Cheng LiuWei-Cheng Liu
7414
7414
$begingroup$
@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
add a comment |
$begingroup$
@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
$begingroup$
@MohammadZuhairKhan No, my question is different. The key point is the parity.
$endgroup$
– Wei-Cheng Liu
Jan 29 at 9:28
$begingroup$
@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
$begingroup$
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
$begingroup$
@MohammadZuhairKhan That is alright. Yes, I will accept the answer below.
$endgroup$
– Wei-Cheng Liu
Jan 30 at 3:08
add a comment |
1 Answer
1
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$begingroup$
$n choose k$ 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
answered Jan 29 at 8:51
Sorin TircSorin Tirc
1,855213
$endgroup$
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1 Answer
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1 Answer
1
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oldest
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$begingroup$
$n choose k$ 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
answered Jan 29 at 8:51
Sorin TircSorin Tirc
1,855213
$endgroup$
add a comment |
$begingroup$
$n choose k$ 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
answered Jan 29 at 8:51
Sorin TircSorin Tirc
1,855213
$endgroup$
add a comment |
$begingroup$
$n choose k$ 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
answered Jan 29 at 8:51
Sorin TircSorin Tirc
1,855213
$endgroup$
$n choose k$ 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
answered Jan 29 at 8:51
Sorin TircSorin Tirc
1,855213
edited Jan 29 at 9:17
answered Jan 29 at 8:51
Sorin TircSorin Tirc
1,855213
answered Jan 29 at 8:51
Sorin TircSorin Tirc
1,855213
answered Jan 29 at 8:51
Sorin TircSorin Tirc
1,855213
1,855213
add a comment |
add a comment |
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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