How to evaluate $ sum_{k=0}^{frac{n}{2}} {nchoose2k}$?
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I would like to evaluate $ sum_{k=0}^{frac{n}{2}} {nchoose2k}$ - but I can't seem to find a simple expression for this sum. Would appreciate any help.
combinatorics summation permutations
asked Nov 23 '18 at 19:25
AlexAlex
70248
$endgroup$
add a comment |
$begingroup$
I would like to evaluate $ sum_{k=0}^{frac{n}{2}} {nchoose2k}$ - but I can't seem to find a simple expression for this sum. Would appreciate any help.
combinatorics summation permutations
asked Nov 23 '18 at 19:25
AlexAlex
70248
$endgroup$
1
$begingroup$
$$(1-1)^n+(1+1)^n=?$$
$endgroup$
– lab bhattacharjee
Nov 23 '18 at 19:27
1
$begingroup$
Is $n{{}}$ even?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:27
add a comment |
$begingroup$
I would like to evaluate $ sum_{k=0}^{frac{n}{2}} {nchoose2k}$ - but I can't seem to find a simple expression for this sum. Would appreciate any help.
combinatorics summation permutations
asked Nov 23 '18 at 19:25
AlexAlex
70248
$endgroup$
I would like to evaluate $ sum_{k=0}^{frac{n}{2}} {nchoose2k}$ - but I can't seem to find a simple expression for this sum. Would appreciate any help.
combinatorics summation permutations
combinatorics summation permutations
asked Nov 23 '18 at 19:25
AlexAlex
70248
asked Nov 23 '18 at 19:25
AlexAlex
70248
asked Nov 23 '18 at 19:25
AlexAlex
70248
asked Nov 23 '18 at 19:25
AlexAlex
70248
asked Nov 23 '18 at 19:25
AlexAlex
70248
70248
1
$begingroup$
$$(1-1)^n+(1+1)^n=?$$
$endgroup$
– lab bhattacharjee
Nov 23 '18 at 19:27
1
$begingroup$
Is $n{{}}$ even?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:27
add a comment |
1
$begingroup$
$$(1-1)^n+(1+1)^n=?$$
$endgroup$
– lab bhattacharjee
Nov 23 '18 at 19:27
1
$begingroup$
Is $n{{}}$ even?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:27
1
1
$begingroup$
$$(1-1)^n+(1+1)^n=?$$
$endgroup$
– lab bhattacharjee
Nov 23 '18 at 19:27
$begingroup$
$$(1-1)^n+(1+1)^n=?$$
$endgroup$
– lab bhattacharjee
Nov 23 '18 at 19:27
1
1
$begingroup$
Is $n{{}}$ even?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:27
$begingroup$
Is $n{{}}$ even?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:27
add a comment |
3 Answers
3
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oldest
votes
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This is the number of subsets of ${1, ldots, n}$ that have even size. If you need a further hint, this question may help.
answered Nov 23 '18 at 19:28
angryavianangryavian
39.5k23280
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add a comment |
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Hint
$binom{n}{r}$ is the number of subsets of $[n]={1,2,3, ldots n}$ of size $r$. So
$sum_{k=0}^{lfloor frac{n}{2} rfloor}binom{n}{2k}$ is the total number of subsets of $[n]$ with even cardinalities and that would be (about :-)) half the total number of subsets of $[n]$.
answered Nov 23 '18 at 19:29
Anurag AAnurag A
25.8k12249
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$begingroup$
Note that small $n$ needs special treatment (hence the smiley)
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– Hagen von Eitzen
Nov 23 '18 at 20:07
add a comment |
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You should probably write that as $$ sum_{k=0}^{[n/2]} binom{n}{2k}$$ otherwise the answer will depend on whether $n$ even of odd. You can equivalently write this as $sum_{k text{ even}, n geq k geq 0 } binom{n}{k}$.
The right way to approach this is to consider $(1-1)^n + (1+1)^n$ as suggested in the comment. Expand this using the binomial theorem.
answered Nov 23 '18 at 19:30
msmmsm
1,248515
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is the number of subsets of ${1, ldots, n}$ that have even size. If you need a further hint, this question may help.
answered Nov 23 '18 at 19:28
angryavianangryavian
39.5k23280
$endgroup$
add a comment |
$begingroup$
This is the number of subsets of ${1, ldots, n}$ that have even size. If you need a further hint, this question may help.
answered Nov 23 '18 at 19:28
angryavianangryavian
39.5k23280
$endgroup$
add a comment |
$begingroup$
This is the number of subsets of ${1, ldots, n}$ that have even size. If you need a further hint, this question may help.
answered Nov 23 '18 at 19:28
angryavianangryavian
39.5k23280
$endgroup$
This is the number of subsets of ${1, ldots, n}$ that have even size. If you need a further hint, this question may help.
answered Nov 23 '18 at 19:28
angryavianangryavian
39.5k23280
answered Nov 23 '18 at 19:28
angryavianangryavian
39.5k23280
answered Nov 23 '18 at 19:28
angryavianangryavian
39.5k23280
answered Nov 23 '18 at 19:28
angryavianangryavian
39.5k23280
39.5k23280
add a comment |
add a comment |
$begingroup$
Hint
$binom{n}{r}$ is the number of subsets of $[n]={1,2,3, ldots n}$ of size $r$. So
$sum_{k=0}^{lfloor frac{n}{2} rfloor}binom{n}{2k}$ is the total number of subsets of $[n]$ with even cardinalities and that would be (about :-)) half the total number of subsets of $[n]$.
answered Nov 23 '18 at 19:29
Anurag AAnurag A
25.8k12249
$endgroup$
$begingroup$
Note that small $n$ needs special treatment (hence the smiley)
$endgroup$
– Hagen von Eitzen
Nov 23 '18 at 20:07
add a comment |
$begingroup$
Hint
$binom{n}{r}$ is the number of subsets of $[n]={1,2,3, ldots n}$ of size $r$. So
$sum_{k=0}^{lfloor frac{n}{2} rfloor}binom{n}{2k}$ is the total number of subsets of $[n]$ with even cardinalities and that would be (about :-)) half the total number of subsets of $[n]$.
answered Nov 23 '18 at 19:29
Anurag AAnurag A
25.8k12249
$endgroup$
$begingroup$
Note that small $n$ needs special treatment (hence the smiley)
$endgroup$
– Hagen von Eitzen
Nov 23 '18 at 20:07
add a comment |
$begingroup$
Hint
$binom{n}{r}$ is the number of subsets of $[n]={1,2,3, ldots n}$ of size $r$. So
$sum_{k=0}^{lfloor frac{n}{2} rfloor}binom{n}{2k}$ is the total number of subsets of $[n]$ with even cardinalities and that would be (about :-)) half the total number of subsets of $[n]$.
answered Nov 23 '18 at 19:29
Anurag AAnurag A
25.8k12249
$endgroup$
Hint
$binom{n}{r}$ is the number of subsets of $[n]={1,2,3, ldots n}$ of size $r$. So
$sum_{k=0}^{lfloor frac{n}{2} rfloor}binom{n}{2k}$ is the total number of subsets of $[n]$ with even cardinalities and that would be (about :-)) half the total number of subsets of $[n]$.
answered Nov 23 '18 at 19:29
Anurag AAnurag A
25.8k12249
edited Nov 23 '18 at 19:36
answered Nov 23 '18 at 19:29
Anurag AAnurag A
25.8k12249
answered Nov 23 '18 at 19:29
Anurag AAnurag A
25.8k12249
answered Nov 23 '18 at 19:29
Anurag AAnurag A
25.8k12249
25.8k12249
$begingroup$
Note that small $n$ needs special treatment (hence the smiley)
$endgroup$
– Hagen von Eitzen
Nov 23 '18 at 20:07
add a comment |
$begingroup$
Note that small $n$ needs special treatment (hence the smiley)
$endgroup$
– Hagen von Eitzen
Nov 23 '18 at 20:07
$begingroup$
Note that small $n$ needs special treatment (hence the smiley)
$endgroup$
– Hagen von Eitzen
Nov 23 '18 at 20:07
$begingroup$
Note that small $n$ needs special treatment (hence the smiley)
$endgroup$
– Hagen von Eitzen
Nov 23 '18 at 20:07
add a comment |
$begingroup$
You should probably write that as $$ sum_{k=0}^{[n/2]} binom{n}{2k}$$ otherwise the answer will depend on whether $n$ even of odd. You can equivalently write this as $sum_{k text{ even}, n geq k geq 0 } binom{n}{k}$.
The right way to approach this is to consider $(1-1)^n + (1+1)^n$ as suggested in the comment. Expand this using the binomial theorem.
answered Nov 23 '18 at 19:30
msmmsm
1,248515
$endgroup$
add a comment |
$begingroup$
You should probably write that as $$ sum_{k=0}^{[n/2]} binom{n}{2k}$$ otherwise the answer will depend on whether $n$ even of odd. You can equivalently write this as $sum_{k text{ even}, n geq k geq 0 } binom{n}{k}$.
The right way to approach this is to consider $(1-1)^n + (1+1)^n$ as suggested in the comment. Expand this using the binomial theorem.
answered Nov 23 '18 at 19:30
msmmsm
1,248515
$endgroup$
add a comment |
$begingroup$
You should probably write that as $$ sum_{k=0}^{[n/2]} binom{n}{2k}$$ otherwise the answer will depend on whether $n$ even of odd. You can equivalently write this as $sum_{k text{ even}, n geq k geq 0 } binom{n}{k}$.
The right way to approach this is to consider $(1-1)^n + (1+1)^n$ as suggested in the comment. Expand this using the binomial theorem.
answered Nov 23 '18 at 19:30
msmmsm
1,248515
$endgroup$
You should probably write that as $$ sum_{k=0}^{[n/2]} binom{n}{2k}$$ otherwise the answer will depend on whether $n$ even of odd. You can equivalently write this as $sum_{k text{ even}, n geq k geq 0 } binom{n}{k}$.
The right way to approach this is to consider $(1-1)^n + (1+1)^n$ as suggested in the comment. Expand this using the binomial theorem.
answered Nov 23 '18 at 19:30
msmmsm
1,248515
answered Nov 23 '18 at 19:30
msmmsm
1,248515
answered Nov 23 '18 at 19:30
msmmsm
1,248515
answered Nov 23 '18 at 19:30
msmmsm
1,248515
1,248515
add a comment |
add a comment |
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1
$begingroup$
$$(1-1)^n+(1+1)^n=?$$
$endgroup$
– lab bhattacharjee
Nov 23 '18 at 19:27
1
$begingroup$
Is $n{{}}$ even?
$endgroup$
– Lord Shark the Unknown
Nov 23 '18 at 19:27