Mixing
Find the sum $S=sumlimits_{k=1}^{n} {4k-1 choose k}$
1
$begingroup$
Find the sum $S=sumlimits_{k=1}^{n} {4k-1 choose k}$
I tried using the Pascal's identity to get $S=sumlimits_{k=1}^{n} {4k choose k}-{4k-1 choose k-1}$ ,but it is not really telescopic.
Any suggestions?
combinatorics algebra-precalculus summation binomial-coefficients
edited Jan 16 at 9:41
rtybase
11k21533
asked Jan 16 at 5:26
Legend KillerLegend Killer
1,6621523
$endgroup$
add a comment |
1
$begingroup$
Find the sum $S=sumlimits_{k=1}^{n} {4k-1 choose k}$
I tried using the Pascal's identity to get $S=sumlimits_{k=1}^{n} {4k choose k}-{4k-1 choose k-1}$ ,but it is not really telescopic.
Any suggestions?
combinatorics algebra-precalculus summation binomial-coefficients
edited Jan 16 at 9:41
rtybase
11k21533
asked Jan 16 at 5:26
Legend KillerLegend Killer
1,6621523
$endgroup$
$begingroup$
You can also have a look at Binomial coefficients#Multisections of sums (current revision) on Wikipedia. Some posts on this site which are (to some extent) similar: Find the value $binom {n}{0} + binom{n}{4} + binom{n}{8} + cdots $, where $n$ is a positive integer., Finding $binom{n}{0} + binom{n}{3} + binom{n}{6} + ldots $ ...
$endgroup$
– Martin Sleziak
Jan 16 at 8:07
$begingroup$
... or How do I count the subsets of a set whose number of elements is divisible by 3? 4? and other posts linked there.
$endgroup$
– Martin Sleziak
Jan 16 at 8:08
$begingroup$
From $color{red}{texttt{Mathematica}}$: $frac{3}{4} left(text{HypergeometricPFQ}left[, _3F_2left(frac{1}{4},frac{1}{2},frac{3}{4};frac{1}{3},frac{2}{3};frac{256}{27}right)right]-1right)-binom{4 n+3}{n+1} text{HypergeometricPFQ}left[, _4F_3left(1,n+frac{5}{4},n+frac{3}{2},n+frac{7}{4};n+frac{4}{3},n+frac{5}{3},n+2;frac{256}{27}right)right]$
$endgroup$
– Felix Marin
Jan 17 at 19:52
$begingroup$
This is amusing: the accepted answer provides the desired value of this sum of $n$ (explicit) terms, as the value of... another sum of $n$ (not-so-explicit) terms. Where is the progress? The OP or one of the upvoters surely will explain this point...
$endgroup$
– Did
Feb 6 at 10:07
add a comment |
1
1
1
0
$begingroup$
Find the sum $S=sumlimits_{k=1}^{n} {4k-1 choose k}$
I tried using the Pascal's identity to get $S=sumlimits_{k=1}^{n} {4k choose k}-{4k-1 choose k-1}$ ,but it is not really telescopic.
Any suggestions?
combinatorics algebra-precalculus summation binomial-coefficients
edited Jan 16 at 9:41
rtybase
11k21533
asked Jan 16 at 5:26
Legend KillerLegend Killer
1,6621523
$endgroup$
Find the sum $S=sumlimits_{k=1}^{n} {4k-1 choose k}$
I tried using the Pascal's identity to get $S=sumlimits_{k=1}^{n} {4k choose k}-{4k-1 choose k-1}$ ,but it is not really telescopic.
Any suggestions?
combinatorics algebra-precalculus summation binomial-coefficients
combinatorics algebra-precalculus summation binomial-coefficients
edited Jan 16 at 9:41
rtybase
11k21533
asked Jan 16 at 5:26
Legend KillerLegend Killer
1,6621523
edited Jan 16 at 9:41
rtybase
11k21533
asked Jan 16 at 5:26
Legend KillerLegend Killer
1,6621523
edited Jan 16 at 9:41
rtybase
11k21533
edited Jan 16 at 9:41
rtybase
11k21533
edited Jan 16 at 9:41
rtybase
11k21533
11k21533
asked Jan 16 at 5:26
Legend KillerLegend Killer
1,6621523
asked Jan 16 at 5:26
Legend KillerLegend Killer
1,6621523
asked Jan 16 at 5:26
Legend KillerLegend Killer
1,6621523
1,6621523
$begingroup$
You can also have a look at Binomial coefficients#Multisections of sums (current revision) on Wikipedia. Some posts on this site which are (to some extent) similar: Find the value $binom {n}{0} + binom{n}{4} + binom{n}{8} + cdots $, where $n$ is a positive integer., Finding $binom{n}{0} + binom{n}{3} + binom{n}{6} + ldots $ ...
$endgroup$
– Martin Sleziak
Jan 16 at 8:07
$begingroup$
... or How do I count the subsets of a set whose number of elements is divisible by 3? 4? and other posts linked there.
$endgroup$
– Martin Sleziak
Jan 16 at 8:08
$begingroup$
From $color{red}{texttt{Mathematica}}$: $frac{3}{4} left(text{HypergeometricPFQ}left[, _3F_2left(frac{1}{4},frac{1}{2},frac{3}{4};frac{1}{3},frac{2}{3};frac{256}{27}right)right]-1right)-binom{4 n+3}{n+1} text{HypergeometricPFQ}left[, _4F_3left(1,n+frac{5}{4},n+frac{3}{2},n+frac{7}{4};n+frac{4}{3},n+frac{5}{3},n+2;frac{256}{27}right)right]$
$endgroup$
– Felix Marin
Jan 17 at 19:52
$begingroup$
This is amusing: the accepted answer provides the desired value of this sum of $n$ (explicit) terms, as the value of... another sum of $n$ (not-so-explicit) terms. Where is the progress? The OP or one of the upvoters surely will explain this point...
$endgroup$
– Did
Feb 6 at 10:07
add a comment |
$begingroup$
You can also have a look at Binomial coefficients#Multisections of sums (current revision) on Wikipedia. Some posts on this site which are (to some extent) similar: Find the value $binom {n}{0} + binom{n}{4} + binom{n}{8} + cdots $, where $n$ is a positive integer., Finding $binom{n}{0} + binom{n}{3} + binom{n}{6} + ldots $ ...
$endgroup$
– Martin Sleziak
Jan 16 at 8:07
$begingroup$
... or How do I count the subsets of a set whose number of elements is divisible by 3? 4? and other posts linked there.
$endgroup$
– Martin Sleziak
Jan 16 at 8:08
$begingroup$
From $color{red}{texttt{Mathematica}}$: $frac{3}{4} left(text{HypergeometricPFQ}left[, _3F_2left(frac{1}{4},frac{1}{2},frac{3}{4};frac{1}{3},frac{2}{3};frac{256}{27}right)right]-1right)-binom{4 n+3}{n+1} text{HypergeometricPFQ}left[, _4F_3left(1,n+frac{5}{4},n+frac{3}{2},n+frac{7}{4};n+frac{4}{3},n+frac{5}{3},n+2;frac{256}{27}right)right]$
$endgroup$
– Felix Marin
Jan 17 at 19:52
$begingroup$
This is amusing: the accepted answer provides the desired value of this sum of $n$ (explicit) terms, as the value of... another sum of $n$ (not-so-explicit) terms. Where is the progress? The OP or one of the upvoters surely will explain this point...
$endgroup$
– Did
Feb 6 at 10:07
$begingroup$
You can also have a look at Binomial coefficients#Multisections of sums (current revision) on Wikipedia. Some posts on this site which are (to some extent) similar: Find the value $binom {n}{0} + binom{n}{4} + binom{n}{8} + cdots $, where $n$ is a positive integer., Finding $binom{n}{0} + binom{n}{3} + binom{n}{6} + ldots $ ...
$endgroup$
– Martin Sleziak
Jan 16 at 8:07
$begingroup$
You can also have a look at Binomial coefficients#Multisections of sums (current revision) on Wikipedia. Some posts on this site which are (to some extent) similar: Find the value $binom {n}{0} + binom{n}{4} + binom{n}{8} + cdots $, where $n$ is a positive integer., Finding $binom{n}{0} + binom{n}{3} + binom{n}{6} + ldots $ ...
$endgroup$
– Martin Sleziak
Jan 16 at 8:07
$begingroup$
... or How do I count the subsets of a set whose number of elements is divisible by 3? 4? and other posts linked there.
$endgroup$
– Martin Sleziak
Jan 16 at 8:08
$begingroup$
... or How do I count the subsets of a set whose number of elements is divisible by 3? 4? and other posts linked there.
$endgroup$
– Martin Sleziak
Jan 16 at 8:08
$begingroup$
From $color{red}{texttt{Mathematica}}$: $frac{3}{4} left(text{HypergeometricPFQ}left[, _3F_2left(frac{1}{4},frac{1}{2},frac{3}{4};frac{1}{3},frac{2}{3};frac{256}{27}right)right]-1right)-binom{4 n+3}{n+1} text{HypergeometricPFQ}left[, _4F_3left(1,n+frac{5}{4},n+frac{3}{2},n+frac{7}{4};n+frac{4}{3},n+frac{5}{3},n+2;frac{256}{27}right)right]$
$endgroup$
– Felix Marin
Jan 17 at 19:52
$begingroup$
From $color{red}{texttt{Mathematica}}$: $frac{3}{4} left(text{HypergeometricPFQ}left[, _3F_2left(frac{1}{4},frac{1}{2},frac{3}{4};frac{1}{3},frac{2}{3};frac{256}{27}right)right]-1right)-binom{4 n+3}{n+1} text{HypergeometricPFQ}left[, _4F_3left(1,n+frac{5}{4},n+frac{3}{2},n+frac{7}{4};n+frac{4}{3},n+frac{5}{3},n+2;frac{256}{27}right)right]$
$endgroup$
– Felix Marin
Jan 17 at 19:52
$begingroup$
This is amusing: the accepted answer provides the desired value of this sum of $n$ (explicit) terms, as the value of... another sum of $n$ (not-so-explicit) terms. Where is the progress? The OP or one of the upvoters surely will explain this point...
$endgroup$
– Did
Feb 6 at 10:07
$begingroup$
This is amusing: the accepted answer provides the desired value of this sum of $n$ (explicit) terms, as the value of... another sum of $n$ (not-so-explicit) terms. Where is the progress? The OP or one of the upvoters surely will explain this point...
$endgroup$
– Did
Feb 6 at 10:07
add a comment |
1 Answer
1
active
oldest
votes
3
$begingroup$
Hint:
The general term of $$(x^a+x^b)^{4k-1}$$ is $$binom{4k-1}rx^{a(4k-1-r)}x^{br}$$
$r=kimplies binom{4k-1}k x^{a(3k-1)+bk}$
To eliminate $k$ in the exponent of $x$
set $3a+b=0iff b=-3a$
WLOG $a=-1,b=?$
We need to find the coefficient of $x$ in $$sum_{k=1}^n(x^{-1}+x^3)^{4k-1}$$ which is a finite Geometric Sequence
answered Jan 16 at 5:39
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075339%2ffind-the-sum-s-sum-limits-k-1n-4k-1-choose-k%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
Hint:
The general term of $$(x^a+x^b)^{4k-1}$$ is $$binom{4k-1}rx^{a(4k-1-r)}x^{br}$$
$r=kimplies binom{4k-1}k x^{a(3k-1)+bk}$
To eliminate $k$ in the exponent of $x$
set $3a+b=0iff b=-3a$
WLOG $a=-1,b=?$
We need to find the coefficient of $x$ in $$sum_{k=1}^n(x^{-1}+x^3)^{4k-1}$$ which is a finite Geometric Sequence
answered Jan 16 at 5:39
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
3
$begingroup$
Hint:
The general term of $$(x^a+x^b)^{4k-1}$$ is $$binom{4k-1}rx^{a(4k-1-r)}x^{br}$$
$r=kimplies binom{4k-1}k x^{a(3k-1)+bk}$
To eliminate $k$ in the exponent of $x$
set $3a+b=0iff b=-3a$
WLOG $a=-1,b=?$
We need to find the coefficient of $x$ in $$sum_{k=1}^n(x^{-1}+x^3)^{4k-1}$$ which is a finite Geometric Sequence
answered Jan 16 at 5:39
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
3
3
3
$begingroup$
Hint:
The general term of $$(x^a+x^b)^{4k-1}$$ is $$binom{4k-1}rx^{a(4k-1-r)}x^{br}$$
$r=kimplies binom{4k-1}k x^{a(3k-1)+bk}$
To eliminate $k$ in the exponent of $x$
set $3a+b=0iff b=-3a$
WLOG $a=-1,b=?$
We need to find the coefficient of $x$ in $$sum_{k=1}^n(x^{-1}+x^3)^{4k-1}$$ which is a finite Geometric Sequence
answered Jan 16 at 5:39
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Hint:
The general term of $$(x^a+x^b)^{4k-1}$$ is $$binom{4k-1}rx^{a(4k-1-r)}x^{br}$$
$r=kimplies binom{4k-1}k x^{a(3k-1)+bk}$
To eliminate $k$ in the exponent of $x$
set $3a+b=0iff b=-3a$
WLOG $a=-1,b=?$
We need to find the coefficient of $x$ in $$sum_{k=1}^n(x^{-1}+x^3)^{4k-1}$$ which is a finite Geometric Sequence
answered Jan 16 at 5:39
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 16 at 5:39
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 16 at 5:39
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 16 at 5:39
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075339%2ffind-the-sum-s-sum-limits-k-1n-4k-1-choose-k%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
You can also have a look at Binomial coefficients#Multisections of sums (current revision) on Wikipedia. Some posts on this site which are (to some extent) similar: Find the value $binom {n}{0} + binom{n}{4} + binom{n}{8} + cdots $, where $n$ is a positive integer., Finding $binom{n}{0} + binom{n}{3} + binom{n}{6} + ldots $ ...
$endgroup$
– Martin Sleziak
Jan 16 at 8:07
$begingroup$
... or How do I count the subsets of a set whose number of elements is divisible by 3? 4? and other posts linked there.
$endgroup$
– Martin Sleziak
Jan 16 at 8:08
$begingroup$
From $color{red}{texttt{Mathematica}}$: $frac{3}{4} left(text{HypergeometricPFQ}left[, _3F_2left(frac{1}{4},frac{1}{2},frac{3}{4};frac{1}{3},frac{2}{3};frac{256}{27}right)right]-1right)-binom{4 n+3}{n+1} text{HypergeometricPFQ}left[, _4F_3left(1,n+frac{5}{4},n+frac{3}{2},n+frac{7}{4};n+frac{4}{3},n+frac{5}{3},n+2;frac{256}{27}right)right]$
$endgroup$
– Felix Marin
Jan 17 at 19:52
$begingroup$
This is amusing: the accepted answer provides the desired value of this sum of $n$ (explicit) terms, as the value of... another sum of $n$ (not-so-explicit) terms. Where is the progress? The OP or one of the upvoters surely will explain this point...
$endgroup$
– Did
Feb 6 at 10:07