Proving by induction of $n$ that $sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}} $
$begingroup$
$$
sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac3{8}$$
RHS-
$$frac{1}{2} - frac1{(n+1)2^{n+1}} = frac3{8}$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_{k=1}^{n+1} frac {k+2}{k(k+1)2^{k+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac{1}{2} - frac1{(n+1)2^{n+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}
$$
and then I broke up $frac {n+3}{(n+1)(n+2)2^{n+2}}$ into
$$frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$
$$=$$
$$frac{2}{(n+1)2^{n+2}} - frac{1}{(n+2)2^{n+2}}$$
$$=$$
$$frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then put it back in with the rest of the equation, bringing me to
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then
$$frac{1}2 -frac{2}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
and
$$frac{1}2 -frac{1}{(n+1)2^{n}} - frac{1}{(n+2)2^{n+2}}$$
$$frac{1}2 -frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$
which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?
$$frac{1}2 -frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$
canceling out $2^{n}$
$$frac{1}2 -frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$
and I'm stuck, please help!
discrete-mathematics induction
edited 18 hours ago
Asaf Karagila♦
307k33439770
asked 23 hours ago
BrownieBrownie
1977
$endgroup$
add a comment |
$begingroup$
$$
sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac3{8}$$
RHS-
$$frac{1}{2} - frac1{(n+1)2^{n+1}} = frac3{8}$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_{k=1}^{n+1} frac {k+2}{k(k+1)2^{k+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac{1}{2} - frac1{(n+1)2^{n+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}
$$
and then I broke up $frac {n+3}{(n+1)(n+2)2^{n+2}}$ into
$$frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$
$$=$$
$$frac{2}{(n+1)2^{n+2}} - frac{1}{(n+2)2^{n+2}}$$
$$=$$
$$frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then put it back in with the rest of the equation, bringing me to
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then
$$frac{1}2 -frac{2}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
and
$$frac{1}2 -frac{1}{(n+1)2^{n}} - frac{1}{(n+2)2^{n+2}}$$
$$frac{1}2 -frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$
which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?
$$frac{1}2 -frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$
canceling out $2^{n}$
$$frac{1}2 -frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$
and I'm stuck, please help!
discrete-mathematics induction
edited 18 hours ago
Asaf Karagila♦
307k33439770
asked 23 hours ago
BrownieBrownie
1977
$endgroup$
add a comment |
$begingroup$
$$
sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac3{8}$$
RHS-
$$frac{1}{2} - frac1{(n+1)2^{n+1}} = frac3{8}$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_{k=1}^{n+1} frac {k+2}{k(k+1)2^{k+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac{1}{2} - frac1{(n+1)2^{n+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}
$$
and then I broke up $frac {n+3}{(n+1)(n+2)2^{n+2}}$ into
$$frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$
$$=$$
$$frac{2}{(n+1)2^{n+2}} - frac{1}{(n+2)2^{n+2}}$$
$$=$$
$$frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then put it back in with the rest of the equation, bringing me to
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then
$$frac{1}2 -frac{2}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
and
$$frac{1}2 -frac{1}{(n+1)2^{n}} - frac{1}{(n+2)2^{n+2}}$$
$$frac{1}2 -frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$
which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?
$$frac{1}2 -frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$
canceling out $2^{n}$
$$frac{1}2 -frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$
and I'm stuck, please help!
discrete-mathematics induction
edited 18 hours ago
Asaf Karagila♦
307k33439770
asked 23 hours ago
BrownieBrownie
1977
$endgroup$
$$
sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
$$
Base Case:
I did $n = 1$, so..
LHS-
$$sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac3{8}$$
RHS-
$$frac{1}{2} - frac1{(n+1)2^{n+1}} = frac3{8}$$
so LHS = RHS
Inductive case-
LHS for $n+1$
$$sum_{k=1}^{n+1} frac {k+2}{k(k+1)2^{k+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}$$
and then I think that you can use inductive hypothesis to change it to the form of
$$
frac{1}{2} - frac1{(n+1)2^{n+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}
$$
and then I broke up $frac {n+3}{(n+1)(n+2)2^{n+2}}$ into
$$frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$
$$=$$
$$frac{2}{(n+1)2^{n+2}} - frac{1}{(n+2)2^{n+2}}$$
$$=$$
$$frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then put it back in with the rest of the equation, bringing me to
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
then
$$frac{1}2 -frac{2}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$
and
$$frac{1}2 -frac{1}{(n+1)2^{n}} - frac{1}{(n+2)2^{n+2}}$$
$$frac{1}2 -frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$
which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?
$$frac{1}2 -frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$
canceling out $2^{n}$
$$frac{1}2 -frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$
and I'm stuck, please help!
discrete-mathematics induction
discrete-mathematics induction
edited 18 hours ago
Asaf Karagila♦
307k33439770
asked 23 hours ago
BrownieBrownie
1977
edited 18 hours ago
Asaf Karagila♦
307k33439770
asked 23 hours ago
BrownieBrownie
1977
edited 18 hours ago
Asaf Karagila♦
307k33439770
edited 18 hours ago
Asaf Karagila♦
307k33439770
edited 18 hours ago
Asaf Karagila♦
307k33439770
307k33439770
asked 23 hours ago
BrownieBrownie
1977
asked 23 hours ago
BrownieBrownie
1977
asked 23 hours ago
BrownieBrownie
1977
1977
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your error is just after the sixth step from the bottom:
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered 23 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
begin{align}
sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
&=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
&=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
&=frac12-frac1{(n+1)2^{n+1}}
end{align}
$$
answered 22 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-that-sum-k-1n-frac-k2kk12k1-fra%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:
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered 23 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:
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered 23 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:
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered 23 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
Your error is just after the sixth step from the bottom:
$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
answered 23 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
answered 23 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
answered 23 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
answered 23 hours ago
John Wayland BalesJohn Wayland Bales
15.1k21238
15.1k21238
add a comment |
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
begin{align}
sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
&=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
&=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
&=frac12-frac1{(n+1)2^{n+1}}
end{align}
$$
answered 22 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
begin{align}
sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
&=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
&=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
&=frac12-frac1{(n+1)2^{n+1}}
end{align}
$$
answered 22 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
add a comment |
$begingroup$
Using a telescoping sum, we get
$$
begin{align}
sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
&=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
&=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
&=frac12-frac1{(n+1)2^{n+1}}
end{align}
$$
answered 22 hours ago
robjohn♦robjohn
270k27312639
$endgroup$
Using a telescoping sum, we get
$$
begin{align}
sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
&=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
&=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
&=frac12-frac1{(n+1)2^{n+1}}
end{align}
$$
answered 22 hours ago
robjohn♦robjohn
270k27312639
answered 22 hours ago
robjohn♦robjohn
270k27312639
answered 22 hours ago
robjohn♦robjohn
270k27312639
answered 22 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-that-sum-k-1n-frac-k2kk12k1-fra%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
