Calculating $sum_{k=0}^{lfloor frac{p}{2} rfloor} binom{p}{k}$
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I'm trying to find the value of: $$sum_{k=0}^{left lfloor frac{p}{2} right rfloor} binom{p}{k}$$ For even and odd $p$, the indication I was given suggests writing it as $$frac{1}{2}left (sum_{k=0}^{left lfloor frac{p}{2} right rfloor} binom{p}{k} + sum_{k=0}^{left lfloor frac{p}{2} right rfloor} binom{p}{p-k} right )$$ But I nothing I did got me anywhere.
summation binomial-coefficients floor-function
asked Nov 17 at 15:54
FuzzyPixelz
337214
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up vote
0
down vote
favorite
I'm trying to find the value of: $$sum_{k=0}^{left lfloor frac{p}{2} right rfloor} binom{p}{k}$$ For even and odd $p$, the indication I was given suggests writing it as $$frac{1}{2}left (sum_{k=0}^{left lfloor frac{p}{2} right rfloor} binom{p}{k} + sum_{k=0}^{left lfloor frac{p}{2} right rfloor} binom{p}{p-k} right )$$ But I nothing I did got me anywhere.
summation binomial-coefficients floor-function
asked Nov 17 at 15:54
FuzzyPixelz
337214
1
Recall that $sum_{k=0}^pbinom{p}{k}=(1+1)^k$.
– Robert Z
Nov 17 at 15:57
Look at Pascal's triangle. The p-th row consists of the values $p choose k$. Maybe you can notice a pattern when you sum 1/2 the values in a row.
– Joel Pereira
Nov 17 at 16:02
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to find the value of: $$sum_{k=0}^{left lfloor frac{p}{2} right rfloor} binom{p}{k}$$ For even and odd $p$, the indication I was given suggests writing it as $$frac{1}{2}left (sum_{k=0}^{left lfloor frac{p}{2} right rfloor} binom{p}{k} + sum_{k=0}^{left lfloor frac{p}{2} right rfloor} binom{p}{p-k} right )$$ But I nothing I did got me anywhere.
summation binomial-coefficients floor-function
asked Nov 17 at 15:54
FuzzyPixelz
337214
I'm trying to find the value of: $$sum_{k=0}^{left lfloor frac{p}{2} right rfloor} binom{p}{k}$$ For even and odd $p$, the indication I was given suggests writing it as $$frac{1}{2}left (sum_{k=0}^{left lfloor frac{p}{2} right rfloor} binom{p}{k} + sum_{k=0}^{left lfloor frac{p}{2} right rfloor} binom{p}{p-k} right )$$ But I nothing I did got me anywhere.
summation binomial-coefficients floor-function
summation binomial-coefficients floor-function
asked Nov 17 at 15:54
FuzzyPixelz
337214
asked Nov 17 at 15:54
FuzzyPixelz
337214
asked Nov 17 at 15:54
FuzzyPixelz
337214
asked Nov 17 at 15:54
FuzzyPixelz
337214
asked Nov 17 at 15:54
FuzzyPixelz
337214
337214
1
Recall that $sum_{k=0}^pbinom{p}{k}=(1+1)^k$.
– Robert Z
Nov 17 at 15:57
Look at Pascal's triangle. The p-th row consists of the values $p choose k$. Maybe you can notice a pattern when you sum 1/2 the values in a row.
– Joel Pereira
Nov 17 at 16:02
add a comment |
1
Recall that $sum_{k=0}^pbinom{p}{k}=(1+1)^k$.
– Robert Z
Nov 17 at 15:57
Look at Pascal's triangle. The p-th row consists of the values $p choose k$. Maybe you can notice a pattern when you sum 1/2 the values in a row.
– Joel Pereira
Nov 17 at 16:02
1
1
Recall that $sum_{k=0}^pbinom{p}{k}=(1+1)^k$.
– Robert Z
Nov 17 at 15:57
Recall that $sum_{k=0}^pbinom{p}{k}=(1+1)^k$.
– Robert Z
Nov 17 at 15:57
Look at Pascal's triangle. The p-th row consists of the values $p choose k$. Maybe you can notice a pattern when you sum 1/2 the values in a row.
– Joel Pereira
Nov 17 at 16:02
Look at Pascal's triangle. The p-th row consists of the values $p choose k$. Maybe you can notice a pattern when you sum 1/2 the values in a row.
– Joel Pereira
Nov 17 at 16:02
add a comment |
1 Answer
1
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0
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You can write
$$
eqalign{
& 2^{,p} = sumlimits_{0, le ,k, le ,p} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
p cr
p - k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,p - leftlfloor {{p over 2}} rightrfloor - 1} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlceil {{p over 2}} rightrceil , le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} = cr
& = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + left( {leftlceil {{p over 2}} rightrceil - leftlfloor {{p over 2}} rightrfloor - 1} right)left( matrix{
p cr
leftlfloor {{p over 2}} rightrfloor cr} right) = cr
& = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} - left( {1 - pbmod 2} right)left( matrix{
p cr
leftlfloor {{p over 2}} rightrfloor cr} right) cr}
$$
answered Nov 18 at 12:13
G Cab
17.1k31237
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You can write
$$
eqalign{
& 2^{,p} = sumlimits_{0, le ,k, le ,p} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
p cr
p - k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,p - leftlfloor {{p over 2}} rightrfloor - 1} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlceil {{p over 2}} rightrceil , le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} = cr
& = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + left( {leftlceil {{p over 2}} rightrceil - leftlfloor {{p over 2}} rightrfloor - 1} right)left( matrix{
p cr
leftlfloor {{p over 2}} rightrfloor cr} right) = cr
& = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} - left( {1 - pbmod 2} right)left( matrix{
p cr
leftlfloor {{p over 2}} rightrfloor cr} right) cr}
$$
answered Nov 18 at 12:13
G Cab
17.1k31237
add a comment |
up vote
0
down vote
You can write
$$
eqalign{
& 2^{,p} = sumlimits_{0, le ,k, le ,p} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
p cr
p - k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,p - leftlfloor {{p over 2}} rightrfloor - 1} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlceil {{p over 2}} rightrceil , le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} = cr
& = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + left( {leftlceil {{p over 2}} rightrceil - leftlfloor {{p over 2}} rightrfloor - 1} right)left( matrix{
p cr
leftlfloor {{p over 2}} rightrfloor cr} right) = cr
& = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} - left( {1 - pbmod 2} right)left( matrix{
p cr
leftlfloor {{p over 2}} rightrfloor cr} right) cr}
$$
answered Nov 18 at 12:13
G Cab
17.1k31237
add a comment |
up vote
0
down vote
up vote
0
down vote
You can write
$$
eqalign{
& 2^{,p} = sumlimits_{0, le ,k, le ,p} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
p cr
p - k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,p - leftlfloor {{p over 2}} rightrfloor - 1} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlceil {{p over 2}} rightrceil , le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} = cr
& = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + left( {leftlceil {{p over 2}} rightrceil - leftlfloor {{p over 2}} rightrfloor - 1} right)left( matrix{
p cr
leftlfloor {{p over 2}} rightrfloor cr} right) = cr
& = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} - left( {1 - pbmod 2} right)left( matrix{
p cr
leftlfloor {{p over 2}} rightrfloor cr} right) cr}
$$
answered Nov 18 at 12:13
G Cab
17.1k31237
You can write
$$
eqalign{
& 2^{,p} = sumlimits_{0, le ,k, le ,p} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
p cr
p - k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,p - leftlfloor {{p over 2}} rightrfloor - 1} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlceil {{p over 2}} rightrceil , le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} = cr
& = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + left( {leftlceil {{p over 2}} rightrceil - leftlfloor {{p over 2}} rightrfloor - 1} right)left( matrix{
p cr
leftlfloor {{p over 2}} rightrfloor cr} right) = cr
& = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} - left( {1 - pbmod 2} right)left( matrix{
p cr
leftlfloor {{p over 2}} rightrfloor cr} right) cr}
$$
answered Nov 18 at 12:13
G Cab
17.1k31237
answered Nov 18 at 12:13
G Cab
17.1k31237
answered Nov 18 at 12:13
G Cab
17.1k31237
answered Nov 18 at 12:13
G Cab
17.1k31237
17.1k31237
add a comment |
add a comment |
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1
Recall that $sum_{k=0}^pbinom{p}{k}=(1+1)^k$.
– Robert Z
Nov 17 at 15:57
Look at Pascal's triangle. The p-th row consists of the values $p choose k$. Maybe you can notice a pattern when you sum 1/2 the values in a row.
– Joel Pereira
Nov 17 at 16:02