Hjdcti

Can a Gentile theist be saved?

Invariance of results when scaling explanatory variables in logistic regression, is there a proof?

Simple image editor tool to draw a simple box/rectangle in an existing image

Partial sums of primes

Did US corporations pay demonstrators in the German demonstrations against article 13?

Can the harmonic series explain the origin of the major scale?

Organic chemistry Iodoform Reaction

The One-Electron Universe postulate is true - what simple change can I make to change the whole universe?

Could solar power be utilized and substitute coal in the 19th century?

What does 사자 in this picture means?

Is a naturally all "male" species possible?

Adding empty element to declared container without declaring type of element

What to do when my ideas aren't chosen, when I strongly disagree with the chosen solution?

How can I raise concerns with a new DM about XP splitting?

Stereotypical names

Hostile work environment after whistle-blowing on coworker and our boss. What do I do?

How can a jailer prevent the Forge Cleric's Artisan's Blessing from being used?

Golf game boilerplate

How will losing mobility of one hand affect my career as a programmer?

Is exact Kanji stroke length important?

Why does this part of the Space Shuttle launch pad seem to be floating in air?

A workplace installs custom certificates on personal devices, can this be used to decrypt HTTPS traffic?

Would it be legal for a US State to ban exports of a natural resource?

Visiting the UK as unmarried couple

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 $sumlimits_k=1^n frac 2^kk leq 2^n$

4

1

$begingroup$

$$
sum_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1
$$

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

share|cite|improve this question

asked 4 hours ago

BrownieBrownie

1927

$endgroup$

add a comment | 

4

1

$begingroup$

$$
sum_k=1^n frac k+2k(k+1)2^k+1 = frac12 - frac1(n+1)2^n+1
$$

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

share|cite|improve this question

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

share|cite|improve this question

asked 4 hours ago

BrownieBrownie

1927

$endgroup$

share|cite|improve this question

asked 4 hours ago

BrownieBrownie

1927

share|cite|improve this question

asked 4 hours ago

BrownieBrownie

1927

asked 4 hours ago

BrownieBrownie

1927

asked 4 hours ago

BrownieBrownie

1927

2 Answers
2

active

oldest

votes

3

$begingroup$

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.

share|cite|improve this answer

answered 4 hours ago

John Wayland BalesJohn Wayland Bales

15.1k21238

$endgroup$

add a comment | 

2

$begingroup$

Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$

share|cite|improve this answer

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
);



);

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

Name

Email

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

Name

Email

Required, but never shown

3

$begingroup$

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.

share|cite|improve this answer

answered 4 hours ago

John Wayland BalesJohn Wayland Bales

15.1k21238

$endgroup$

add a comment | 

3

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered 4 hours ago

John Wayland BalesJohn Wayland Bales

15.1k21238

$endgroup$

share|cite|improve this answer

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

2

$begingroup$

Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$

share|cite|improve this answer

answered 3 hours ago

robjohn♦robjohn

270k27312639

$endgroup$

add a comment | 

2

$begingroup$

Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$

share|cite|improve this answer

answered 3 hours ago

robjohn♦robjohn

270k27312639

$endgroup$

add a comment | 

$begingroup$

Using a telescoping sum, we get
$$
beginalign
sum_k=1^nfrack+2k(k+1)2^k+1
&=sum_k=1^nleft(frac1k2^k-frac1(k+1)2^k+1right)\
&=sum_k=1^nfrac1k2^k-sum_k=2^n+1frac1k2^k\
&=frac12-frac1(n+1)2^n+1
endalign
$$

share|cite|improve this answer

answered 3 hours ago

robjohn♦robjohn

270k27312639

$endgroup$

share|cite|improve this answer

answered 3 hours ago

robjohn♦robjohn

270k27312639

answered 3 hours ago

robjohn♦robjohn

270k27312639

answered 3 hours ago

robjohn♦robjohn

270k27312639