Mixing
Solve recursion
$begingroup$
$a_n=n^2times a_{(n-1)}$, for $ n>0, a_0=1$
Whats the correct formula to use in this case?
I solved problem with recursion before, but they had '$+$' instead of $'times'$ so there is following formula to simply solve them:
$ar^n+bnr^n$.
Is there something similar, but for multiplication instead?
discrete-mathematics
edited Jan 26 at 23:29
abc...
3,237738
asked Jan 26 at 23:27
GorossoGorosso
315
$endgroup$
add a comment |
$begingroup$
$a_n=n^2times a_{(n-1)}$, for $ n>0, a_0=1$
Whats the correct formula to use in this case?
I solved problem with recursion before, but they had '$+$' instead of $'times'$ so there is following formula to simply solve them:
$ar^n+bnr^n$.
Is there something similar, but for multiplication instead?
discrete-mathematics
edited Jan 26 at 23:29
abc...
3,237738
asked Jan 26 at 23:27
GorossoGorosso
315
$endgroup$
1
$begingroup$
Compute the first terms for $sqrt{a_n}$.
$endgroup$
– Mindlack
Jan 26 at 23:29
add a comment |
$begingroup$
$a_n=n^2times a_{(n-1)}$, for $ n>0, a_0=1$
Whats the correct formula to use in this case?
I solved problem with recursion before, but they had '$+$' instead of $'times'$ so there is following formula to simply solve them:
$ar^n+bnr^n$.
Is there something similar, but for multiplication instead?
discrete-mathematics
edited Jan 26 at 23:29
abc...
3,237738
asked Jan 26 at 23:27
GorossoGorosso
315
$endgroup$
$a_n=n^2times a_{(n-1)}$, for $ n>0, a_0=1$
Whats the correct formula to use in this case?
I solved problem with recursion before, but they had '$+$' instead of $'times'$ so there is following formula to simply solve them:
$ar^n+bnr^n$.
Is there something similar, but for multiplication instead?
discrete-mathematics
discrete-mathematics
edited Jan 26 at 23:29
abc...
3,237738
asked Jan 26 at 23:27
GorossoGorosso
315
edited Jan 26 at 23:29
abc...
3,237738
asked Jan 26 at 23:27
GorossoGorosso
315
edited Jan 26 at 23:29
abc...
3,237738
edited Jan 26 at 23:29
abc...
3,237738
edited Jan 26 at 23:29
abc...
3,237738
3,237738
asked Jan 26 at 23:27
GorossoGorosso
315
asked Jan 26 at 23:27
GorossoGorosso
315
asked Jan 26 at 23:27
GorossoGorosso
315
315
1
$begingroup$
Compute the first terms for $sqrt{a_n}$.
$endgroup$
– Mindlack
Jan 26 at 23:29
add a comment |
1
$begingroup$
Compute the first terms for $sqrt{a_n}$.
$endgroup$
– Mindlack
Jan 26 at 23:29
1
1
$begingroup$
Compute the first terms for $sqrt{a_n}$.
$endgroup$
– Mindlack
Jan 26 at 23:29
$begingroup$
Compute the first terms for $sqrt{a_n}$.
$endgroup$
– Mindlack
Jan 26 at 23:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$a_n=(n!)^2$ using induction. Try to evaluate the first few terms on your own.
answered Jan 26 at 23:30
abc...abc...
3,237738
$endgroup$
$begingroup$
This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
$endgroup$
– Klaus
Jan 26 at 23:37
add a comment |
$begingroup$
Multiply these lines together:
begin{align}
a_n&=n^2times a_{n-1}\
a_{n-1}&=(n-1)^2times a_{n-2}\
&dots\
a_2&=2^2times a_{1}\
a_1&=1^2times a_{0}\
end{align}
After canceling, we get
$$
a_n = 1^2 cdot 2^2 cdots n^2 a_0 = (1 cdot 2 cdots n)^2 = (n!)^2
$$
answered Jan 27 at 0:37
lhflhf
166k11172402
$endgroup$
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
});
}
});
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%2f3088920%2fsolve-recursion%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$a_n=(n!)^2$ using induction. Try to evaluate the first few terms on your own.
answered Jan 26 at 23:30
abc...abc...
3,237738
$endgroup$
$begingroup$
This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
$endgroup$
– Klaus
Jan 26 at 23:37
add a comment |
$begingroup$
$a_n=(n!)^2$ using induction. Try to evaluate the first few terms on your own.
answered Jan 26 at 23:30
abc...abc...
3,237738
$endgroup$
$begingroup$
This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
$endgroup$
– Klaus
Jan 26 at 23:37
add a comment |
$begingroup$
$a_n=(n!)^2$ using induction. Try to evaluate the first few terms on your own.
answered Jan 26 at 23:30
abc...abc...
3,237738
$endgroup$
$a_n=(n!)^2$ using induction. Try to evaluate the first few terms on your own.
answered Jan 26 at 23:30
abc...abc...
3,237738
answered Jan 26 at 23:30
abc...abc...
3,237738
answered Jan 26 at 23:30
abc...abc...
3,237738
answered Jan 26 at 23:30
abc...abc...
3,237738
3,237738
$begingroup$
This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
$endgroup$
– Klaus
Jan 26 at 23:37
add a comment |
$begingroup$
This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
$endgroup$
– Klaus
Jan 26 at 23:37
$begingroup$
This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
$endgroup$
– Klaus
Jan 26 at 23:37
$begingroup$
This is the only reasonable answer, but it shows that the problem is somewhat stupid as $n!$ is defined literally like that, i.e. $n! = n cdot (n-1) cdot ldots cdot 2 cdot 1$...
$endgroup$
– Klaus
Jan 26 at 23:37
add a comment |
$begingroup$
Multiply these lines together:
begin{align}
a_n&=n^2times a_{n-1}\
a_{n-1}&=(n-1)^2times a_{n-2}\
&dots\
a_2&=2^2times a_{1}\
a_1&=1^2times a_{0}\
end{align}
After canceling, we get
$$
a_n = 1^2 cdot 2^2 cdots n^2 a_0 = (1 cdot 2 cdots n)^2 = (n!)^2
$$
answered Jan 27 at 0:37
lhflhf
166k11172402
$endgroup$
add a comment |
$begingroup$
Multiply these lines together:
begin{align}
a_n&=n^2times a_{n-1}\
a_{n-1}&=(n-1)^2times a_{n-2}\
&dots\
a_2&=2^2times a_{1}\
a_1&=1^2times a_{0}\
end{align}
After canceling, we get
$$
a_n = 1^2 cdot 2^2 cdots n^2 a_0 = (1 cdot 2 cdots n)^2 = (n!)^2
$$
answered Jan 27 at 0:37
lhflhf
166k11172402
$endgroup$
add a comment |
$begingroup$
Multiply these lines together:
begin{align}
a_n&=n^2times a_{n-1}\
a_{n-1}&=(n-1)^2times a_{n-2}\
&dots\
a_2&=2^2times a_{1}\
a_1&=1^2times a_{0}\
end{align}
After canceling, we get
$$
a_n = 1^2 cdot 2^2 cdots n^2 a_0 = (1 cdot 2 cdots n)^2 = (n!)^2
$$
answered Jan 27 at 0:37
lhflhf
166k11172402
$endgroup$
Multiply these lines together:
begin{align}
a_n&=n^2times a_{n-1}\
a_{n-1}&=(n-1)^2times a_{n-2}\
&dots\
a_2&=2^2times a_{1}\
a_1&=1^2times a_{0}\
end{align}
After canceling, we get
$$
a_n = 1^2 cdot 2^2 cdots n^2 a_0 = (1 cdot 2 cdots n)^2 = (n!)^2
$$
answered Jan 27 at 0:37
lhflhf
166k11172402
answered Jan 27 at 0:37
lhflhf
166k11172402
answered Jan 27 at 0:37
lhflhf
166k11172402
answered Jan 27 at 0:37
lhflhf
166k11172402
166k11172402
add a comment |
add a comment |
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.
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%2f3088920%2fsolve-recursion%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
1
$begingroup$
Compute the first terms for $sqrt{a_n}$.
$endgroup$
– Mindlack
Jan 26 at 23:29