Proving by induction of n. Is this correct until this point?prove inequality by induction — Discrete mathProve $25^n>6^n$ using inductionTrying to simplify an expression for an induction proof.Induction on summation inequality stuck on Induction stepProve by Induction: Summation of Factorial (n! * n)Prove that $n! > n^3$ for every integer $n ge 6$ using inductionProving by induction on $n$ that $sum limits_k=1^n (k+1)2^k = n2^n+1 $5. Prove by induction on $n$ that $sumlimits_k=1^n frac kk+1 leq n - frac1n+1$Prove by induction on n that $sumlimits_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1$Prove by induction on n that $sumlimits_k=1^n frac 2^kk leq 2^n$
How do I rename a LINUX host without needing to reboot for the rename to take effect?
Does "Dominei" mean something?
Superhero words!
How do I repair my stair bannister?
Is there an wasy way to program in Tikz something like the one in the image?
"lassen" in meaning "sich fassen"
Is there an Impartial Brexit Deal comparison site?
Is there a problem with hiding "forgot password" until it's needed?
No idea how to draw this using tikz
Can I rely on these GitHub repository files?
Teaching indefinite integrals that require special-casing
What if somebody invests in my application?
Simulating a probability of 1 of 2^N with less than N random bits
Simple recursive Sudoku solver
Reply ‘no position’ while the job posting is still there (‘HiWi’ position in Germany)
My boss asked me to take a one-day class, then signs it up as a day off
Resetting two CD4017 counters simultaneously, only one resets
What do you call the infoboxes with text and sometimes images on the side of a page we find in textbooks?
How to color a zone in Tikz
What to do when my ideas aren't chosen, when I strongly disagree with the chosen solution?
Books on the History of math research at European universities
Is there enough fresh water in the world to eradicate the drinking water crisis?
Pronouncing Homer as in modern Greek
Golf game boilerplate
Proving by induction of n. Is this correct until this point?
prove inequality by induction — Discrete mathProve $25^n>6^n$ using inductionTrying to simplify an expression for an induction proof.Induction on summation inequality stuck on Induction stepProve by Induction: Summation of Factorial (n! * n)Prove that $n! > n^3$ for every integer $n ge 6$ using inductionProving by induction on $n$ that $sum limits_k=1^n (k+1)2^k = n2^n+1 $5. Prove by induction on $n$ that $sumlimits_k=1^n frac kk+1 leq n - frac1n+1$Prove by induction on n that $sumlimits_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1$Prove by induction on n that $sumlimits_k=1^n frac 2^kk leq 2^n$
$begingroup$
$$
sum_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_k=1^n frac k+2k(k+1)2^k+1 = frac38$$
RHS-
$$frac12 - frac1(n+1)2^n+1 = frac38$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_k=1^n+1 frac k+2k(k+1)2^k+1 +frac n+3(n+1)(n+2)2^n+2$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac12 - frac1(n+1)2^n+1 +frac n+3(n+1)(n+2)2^n+2
$$
and then I broke up $frac n+3(n+1)(n+2)2^n+2$ into
$$frac2(n+2)-(n+1)(n+1)(n+2)2^n+2$$
$$=$$
$$frac2(n+1)2^n+2 - frac1(n+2)2^n+2$$
$$=$$
$$frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then put it back in with the rest of the equation, bringing me to
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then
$$frac12 -frac2(n+1)2^n+1 - frac1(n+2)2^n+2$$
and
$$frac12 -frac1(n+1)2^n - frac1(n+2)2^n+2$$
$$frac12 -frac(n+2)2^n+2 - (n+1)2^n(n+1)(n+2)2^2n+2 $$
which I think simplifies down to this after factoring out a $2^n$ from the numerator?
$$frac12 -frac2^n((n+2)2^2 - (n+1))(n+1)(n+2)2^2n+2 $$
canceling out $2^n$
$$frac12 -frac(3n-7)(n+1)(n+2)2^n+2 $$
and I'm stuck, please help!
discrete-mathematics induction
asked 4 hours ago
BrownieBrownie
1927
$endgroup$
add a comment |
$begingroup$
$$
sum_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_k=1^n frac k+2k(k+1)2^k+1 = frac38$$
RHS-
$$frac12 - frac1(n+1)2^n+1 = frac38$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_k=1^n+1 frac k+2k(k+1)2^k+1 +frac n+3(n+1)(n+2)2^n+2$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac12 - frac1(n+1)2^n+1 +frac n+3(n+1)(n+2)2^n+2
$$
and then I broke up $frac n+3(n+1)(n+2)2^n+2$ into
$$frac2(n+2)-(n+1)(n+1)(n+2)2^n+2$$
$$=$$
$$frac2(n+1)2^n+2 - frac1(n+2)2^n+2$$
$$=$$
$$frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then put it back in with the rest of the equation, bringing me to
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then
$$frac12 -frac2(n+1)2^n+1 - frac1(n+2)2^n+2$$
and
$$frac12 -frac1(n+1)2^n - frac1(n+2)2^n+2$$
$$frac12 -frac(n+2)2^n+2 - (n+1)2^n(n+1)(n+2)2^2n+2 $$
which I think simplifies down to this after factoring out a $2^n$ from the numerator?
$$frac12 -frac2^n((n+2)2^2 - (n+1))(n+1)(n+2)2^2n+2 $$
canceling out $2^n$
$$frac12 -frac(3n-7)(n+1)(n+2)2^n+2 $$
and I'm stuck, please help!
discrete-mathematics induction
asked 4 hours ago
BrownieBrownie
1927
$endgroup$
add a comment |
$begingroup$
$$
sum_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_k=1^n frac k+2k(k+1)2^k+1 = frac38$$
RHS-
$$frac12 - frac1(n+1)2^n+1 = frac38$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_k=1^n+1 frac k+2k(k+1)2^k+1 +frac n+3(n+1)(n+2)2^n+2$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac12 - frac1(n+1)2^n+1 +frac n+3(n+1)(n+2)2^n+2
$$
and then I broke up $frac n+3(n+1)(n+2)2^n+2$ into
$$frac2(n+2)-(n+1)(n+1)(n+2)2^n+2$$
$$=$$
$$frac2(n+1)2^n+2 - frac1(n+2)2^n+2$$
$$=$$
$$frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then put it back in with the rest of the equation, bringing me to
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then
$$frac12 -frac2(n+1)2^n+1 - frac1(n+2)2^n+2$$
and
$$frac12 -frac1(n+1)2^n - frac1(n+2)2^n+2$$
$$frac12 -frac(n+2)2^n+2 - (n+1)2^n(n+1)(n+2)2^2n+2 $$
which I think simplifies down to this after factoring out a $2^n$ from the numerator?
$$frac12 -frac2^n((n+2)2^2 - (n+1))(n+1)(n+2)2^2n+2 $$
canceling out $2^n$
$$frac12 -frac(3n-7)(n+1)(n+2)2^n+2 $$
and I'm stuck, please help!
discrete-mathematics induction
asked 4 hours ago
BrownieBrownie
1927
$endgroup$
$$
sum_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_k=1^n frac k+2k(k+1)2^k+1 = frac38$$
RHS-
$$frac12 - frac1(n+1)2^n+1 = frac38$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_k=1^n+1 frac k+2k(k+1)2^k+1 +frac n+3(n+1)(n+2)2^n+2$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac12 - frac1(n+1)2^n+1 +frac n+3(n+1)(n+2)2^n+2
$$
and then I broke up $frac n+3(n+1)(n+2)2^n+2$ into
$$frac2(n+2)-(n+1)(n+1)(n+2)2^n+2$$
$$=$$
$$frac2(n+1)2^n+2 - frac1(n+2)2^n+2$$
$$=$$
$$frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then put it back in with the rest of the equation, bringing me to
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2$$
then
$$frac12 -frac2(n+1)2^n+1 - frac1(n+2)2^n+2$$
and
$$frac12 -frac1(n+1)2^n - frac1(n+2)2^n+2$$
$$frac12 -frac(n+2)2^n+2 - (n+1)2^n(n+1)(n+2)2^2n+2 $$
which I think simplifies down to this after factoring out a $2^n$ from the numerator?
$$frac12 -frac2^n((n+2)2^2 - (n+1))(n+1)(n+2)2^2n+2 $$
canceling out $2^n$
$$frac12 -frac(3n-7)(n+1)(n+2)2^n+2 $$
and I'm stuck, please help!
discrete-mathematics induction
discrete-mathematics induction
asked 4 hours ago
BrownieBrownie
1927
asked 4 hours ago
BrownieBrownie
1927
asked 4 hours ago
BrownieBrownie
1927
asked 4 hours ago
BrownieBrownie
1927
asked 4 hours ago
BrownieBrownie
1927
1927
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your error is just after the sixth step from the bottom:
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2=frac12 -frac1(n+2)2^n+2$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$
answered 3 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
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%2f3162553%2fproving-by-induction-of-n-is-this-correct-until-this-point%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
$begingroup$
Your error is just after the sixth step from the bottom:
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2=frac12 -frac1(n+2)2^n+2$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
Your error is just after the sixth step from the bottom:
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2=frac12 -frac1(n+2)2^n+2$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
Your error is just after the sixth step from the bottom:
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2=frac12 -frac1(n+2)2^n+2$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
Your error is just after the sixth step from the bottom:
$$frac12 -frac 1(n+1)2^n+1 +frac1(n+1)2^n+1 - frac1(n+2)2^n+2=frac12 -frac1(n+2)2^n+2$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
answered 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
answered 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
answered 4 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
15.1k21238
add a comment |
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$
answered 3 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$
answered 3 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$
answered 3 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$
answered 3 hours ago
robjohn♦robjohn
270k27312639
answered 3 hours ago
robjohn♦robjohn
270k27312639
answered 3 hours ago
robjohn♦robjohn
270k27312639
answered 3 hours ago
robjohn♦robjohn
270k27312639
270k27312639
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%2f3162553%2fproving-by-induction-of-n-is-this-correct-until-this-point%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
