Mixing
Prove by induction $(n+1) + (n + 2) +dots + (2n - 1 ) + (2n) = frac{n(3n + 1)}{2}$, $n geq 1$
$begingroup$
Prove by induction that
$(n+1) + (n + 2) +dots + (2n - 1 ) + (2n) = dfrac{n(3n + 1)}{2}$
for $n geq 1$.
I tried to add $(2(k+1)-1) + (2(k+1))$ to both sides and got stuck.
Any other ideas?
discrete-mathematics summation induction arithmetic-progressions
edited Jan 17 at 14:13
Robert Z
98.6k1068139
asked Jan 17 at 14:04
VLadosVLados
1
$endgroup$
add a comment |
$begingroup$
Prove by induction that
$(n+1) + (n + 2) +dots + (2n - 1 ) + (2n) = dfrac{n(3n + 1)}{2}$
for $n geq 1$.
I tried to add $(2(k+1)-1) + (2(k+1))$ to both sides and got stuck.
Any other ideas?
discrete-mathematics summation induction arithmetic-progressions
edited Jan 17 at 14:13
Robert Z
98.6k1068139
asked Jan 17 at 14:04
VLadosVLados
1
$endgroup$
$begingroup$
Don't forget that the first term changes as well.
$endgroup$
– lulu
Jan 17 at 14:10
add a comment |
$begingroup$
Prove by induction that
$(n+1) + (n + 2) +dots + (2n - 1 ) + (2n) = dfrac{n(3n + 1)}{2}$
for $n geq 1$.
I tried to add $(2(k+1)-1) + (2(k+1))$ to both sides and got stuck.
Any other ideas?
discrete-mathematics summation induction arithmetic-progressions
edited Jan 17 at 14:13
Robert Z
98.6k1068139
asked Jan 17 at 14:04
VLadosVLados
1
$endgroup$
Prove by induction that
$(n+1) + (n + 2) +dots + (2n - 1 ) + (2n) = dfrac{n(3n + 1)}{2}$
for $n geq 1$.
I tried to add $(2(k+1)-1) + (2(k+1))$ to both sides and got stuck.
Any other ideas?
discrete-mathematics summation induction arithmetic-progressions
discrete-mathematics summation induction arithmetic-progressions
edited Jan 17 at 14:13
Robert Z
98.6k1068139
asked Jan 17 at 14:04
VLadosVLados
1
edited Jan 17 at 14:13
Robert Z
98.6k1068139
asked Jan 17 at 14:04
VLadosVLados
1
edited Jan 17 at 14:13
Robert Z
98.6k1068139
edited Jan 17 at 14:13
Robert Z
98.6k1068139
edited Jan 17 at 14:13
Robert Z
98.6k1068139
98.6k1068139
asked Jan 17 at 14:04
VLadosVLados
1
asked Jan 17 at 14:04
VLadosVLados
1
asked Jan 17 at 14:04
VLadosVLados
1
1
$begingroup$
Don't forget that the first term changes as well.
$endgroup$
– lulu
Jan 17 at 14:10
add a comment |
$begingroup$
Don't forget that the first term changes as well.
$endgroup$
– lulu
Jan 17 at 14:10
$begingroup$
Don't forget that the first term changes as well.
$endgroup$
– lulu
Jan 17 at 14:10
$begingroup$
Don't forget that the first term changes as well.
$endgroup$
– lulu
Jan 17 at 14:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, in the inductive step we start from
$$(n+1) + (n + 2) +dots+ (2n – 1 ) + (2n) = n(3n + 1)/2.$$
We add $ (2(n+1)-1) + (2(n+1))$ to both sides
AND we subtract $n+1$ from both sides.
Then we verify that the right hand side is equal to
$$ (n+1)(3(n+1) + 1)/2.$$
Can you take it form here?
answered Jan 17 at 14:09
Robert ZRobert Z
98.6k1068139
$endgroup$
$begingroup$
Yes, I was editing was answer. Thanks!
$endgroup$
– Robert Z
Jan 17 at 14:11
$begingroup$
Great!! Thank you
$endgroup$
– VLados
Jan 17 at 14:11
add a comment |
$begingroup$
If you want to be sneaky, show the easier $s(n) := sumlimits_{k=1}^n k = frac{n(n+1)}{2}$ and then subtract $s(n)$ from $s(2n)$.
answered Jan 17 at 14:33
KlausKlaus
1,8179
$endgroup$
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%2f3077012%2fprove-by-induction-n1-n-2-dots-2n-1-2n-fracn3n-1%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
$begingroup$
Yes, in the inductive step we start from
$$(n+1) + (n + 2) +dots+ (2n – 1 ) + (2n) = n(3n + 1)/2.$$
We add $ (2(n+1)-1) + (2(n+1))$ to both sides
AND we subtract $n+1$ from both sides.
Then we verify that the right hand side is equal to
$$ (n+1)(3(n+1) + 1)/2.$$
Can you take it form here?
answered Jan 17 at 14:09
Robert ZRobert Z
98.6k1068139
$endgroup$
$begingroup$
Yes, I was editing was answer. Thanks!
$endgroup$
– Robert Z
Jan 17 at 14:11
$begingroup$
Great!! Thank you
$endgroup$
– VLados
Jan 17 at 14:11
add a comment |
$begingroup$
Yes, in the inductive step we start from
$$(n+1) + (n + 2) +dots+ (2n – 1 ) + (2n) = n(3n + 1)/2.$$
We add $ (2(n+1)-1) + (2(n+1))$ to both sides
AND we subtract $n+1$ from both sides.
Then we verify that the right hand side is equal to
$$ (n+1)(3(n+1) + 1)/2.$$
Can you take it form here?
answered Jan 17 at 14:09
Robert ZRobert Z
98.6k1068139
$endgroup$
$begingroup$
Yes, I was editing was answer. Thanks!
$endgroup$
– Robert Z
Jan 17 at 14:11
$begingroup$
Great!! Thank you
$endgroup$
– VLados
Jan 17 at 14:11
add a comment |
$begingroup$
Yes, in the inductive step we start from
$$(n+1) + (n + 2) +dots+ (2n – 1 ) + (2n) = n(3n + 1)/2.$$
We add $ (2(n+1)-1) + (2(n+1))$ to both sides
AND we subtract $n+1$ from both sides.
Then we verify that the right hand side is equal to
$$ (n+1)(3(n+1) + 1)/2.$$
Can you take it form here?
answered Jan 17 at 14:09
Robert ZRobert Z
98.6k1068139
$endgroup$
Yes, in the inductive step we start from
$$(n+1) + (n + 2) +dots+ (2n – 1 ) + (2n) = n(3n + 1)/2.$$
We add $ (2(n+1)-1) + (2(n+1))$ to both sides
AND we subtract $n+1$ from both sides.
Then we verify that the right hand side is equal to
$$ (n+1)(3(n+1) + 1)/2.$$
Can you take it form here?
answered Jan 17 at 14:09
Robert ZRobert Z
98.6k1068139
edited Jan 17 at 14:12
answered Jan 17 at 14:09
Robert ZRobert Z
98.6k1068139
answered Jan 17 at 14:09
Robert ZRobert Z
98.6k1068139
answered Jan 17 at 14:09
Robert ZRobert Z
98.6k1068139
98.6k1068139
$begingroup$
Yes, I was editing was answer. Thanks!
$endgroup$
– Robert Z
Jan 17 at 14:11
$begingroup$
Great!! Thank you
$endgroup$
– VLados
Jan 17 at 14:11
add a comment |
$begingroup$
Yes, I was editing was answer. Thanks!
$endgroup$
– Robert Z
Jan 17 at 14:11
$begingroup$
Great!! Thank you
$endgroup$
– VLados
Jan 17 at 14:11
$begingroup$
Yes, I was editing was answer. Thanks!
$endgroup$
– Robert Z
Jan 17 at 14:11
$begingroup$
Yes, I was editing was answer. Thanks!
$endgroup$
– Robert Z
Jan 17 at 14:11
$begingroup$
Great!! Thank you
$endgroup$
– VLados
Jan 17 at 14:11
$begingroup$
Great!! Thank you
$endgroup$
– VLados
Jan 17 at 14:11
add a comment |
$begingroup$
If you want to be sneaky, show the easier $s(n) := sumlimits_{k=1}^n k = frac{n(n+1)}{2}$ and then subtract $s(n)$ from $s(2n)$.
answered Jan 17 at 14:33
KlausKlaus
1,8179
$endgroup$
add a comment |
$begingroup$
If you want to be sneaky, show the easier $s(n) := sumlimits_{k=1}^n k = frac{n(n+1)}{2}$ and then subtract $s(n)$ from $s(2n)$.
answered Jan 17 at 14:33
KlausKlaus
1,8179
$endgroup$
add a comment |
$begingroup$
If you want to be sneaky, show the easier $s(n) := sumlimits_{k=1}^n k = frac{n(n+1)}{2}$ and then subtract $s(n)$ from $s(2n)$.
answered Jan 17 at 14:33
KlausKlaus
1,8179
$endgroup$
If you want to be sneaky, show the easier $s(n) := sumlimits_{k=1}^n k = frac{n(n+1)}{2}$ and then subtract $s(n)$ from $s(2n)$.
answered Jan 17 at 14:33
KlausKlaus
1,8179
answered Jan 17 at 14:33
KlausKlaus
1,8179
answered Jan 17 at 14:33
KlausKlaus
1,8179
answered Jan 17 at 14:33
KlausKlaus
1,8179
1,8179
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%2f3077012%2fprove-by-induction-n1-n-2-dots-2n-1-2n-fracn3n-1%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
$begingroup$
Don't forget that the first term changes as well.
$endgroup$
– lulu
Jan 17 at 14:10