Mixing
How to show that $4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}$ equal to...
$begingroup$
$$4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}$$
How can one algebraically show show that they are equal by starting at the left side? How would one do it without expanding the sum but instead by manipulating the sums directly?
Background: This summation is part of some calculations surrounding Simpson's Rule.
algebra-precalculus summation
asked Jan 20 at 15:51
GambitSquaredGambitSquared
1,1901138
$endgroup$
add a comment |
$begingroup$
$$4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}$$
How can one algebraically show show that they are equal by starting at the left side? How would one do it without expanding the sum but instead by manipulating the sums directly?
Background: This summation is part of some calculations surrounding Simpson's Rule.
algebra-precalculus summation
asked Jan 20 at 15:51
GambitSquaredGambitSquared
1,1901138
$endgroup$
add a comment |
$begingroup$
$$4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}$$
How can one algebraically show show that they are equal by starting at the left side? How would one do it without expanding the sum but instead by manipulating the sums directly?
Background: This summation is part of some calculations surrounding Simpson's Rule.
algebra-precalculus summation
asked Jan 20 at 15:51
GambitSquaredGambitSquared
1,1901138
$endgroup$
$$4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}$$
How can one algebraically show show that they are equal by starting at the left side? How would one do it without expanding the sum but instead by manipulating the sums directly?
Background: This summation is part of some calculations surrounding Simpson's Rule.
algebra-precalculus summation
algebra-precalculus summation
asked Jan 20 at 15:51
GambitSquaredGambitSquared
1,1901138
asked Jan 20 at 15:51
GambitSquaredGambitSquared
1,1901138
edited Jan 21 at 0:08
GambitSquared
asked Jan 20 at 15:51
GambitSquaredGambitSquared
1,1901138
asked Jan 20 at 15:51
GambitSquaredGambitSquared
1,1901138
asked Jan 20 at 15:51
GambitSquaredGambitSquared
1,1901138
1,1901138
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Maybe the multiple use of $j$ as index of summation is a bit confusing. Just note that
$$sum_{k=1}^{2n-1} y_k = underbrace{sum_{stackrel{k=1}{k: odd}}^{2n-1} y_k}_{=sum_{j=1}^{n} y_{2j-1} } + underbrace{sum_{stackrel{k=1}{k: even}}^{2n-1} y_k}_{=sum_{j=1}^{n-1} y_{2j}}$$
Now, you get immediately
$$4sum_{k=1}^{2n-1} y_k = 4 sum_{j=1}^{n} y_{2j-1} + underbrace{4}_{=2+2}sum_{j=1}^{n-1} y_{2j}Leftrightarrow$$
$$4sum_{k=1}^{2n-1} y_k - 2sum_{j=1}^{n-1} y_{2j} = 4 sum_{j=1}^{n} y_{2j-1} + 2sum_{j=1}^{n-1} y_{2j}$$
If you like you may now replace index $k$ by $j$ in the left-most sum.
answered Jan 21 at 11:50
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
(NB - this is for illustration only)
$$4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}$$
As an illustration, take $n=4$.
LHS = $$begin{align}
4big(&y_1&&+y_3&&+y_5&&+y_7big)\
+2big(&&y_2&&+y_4&&+y_6&;;;;;;big)\
=;;;&4y_1&+2y_2&+4y_3&+2y_4&+4y_5&+2y_6&+4y_7end{align}$$
RHS=
$$begin{align}
4big(&y_1&+y_2&+y_3&+y_4&+y_5&y_6&+y_7big)\
-2big(&&y_2&&+y_4&&+y_6&;;;;;;big)\
=;;;&4y_1&+2y_2&+4y_3&+2y_4&+4y_5&+2y_6&+4y_7end{align}$$
Hence LHS=RHS.
Formal Proof
$$begin{align}
text{LHS}&=4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}\
&=underbrace{4sum_{j=1}^ny_{2j-1}color{red}{+4sum_{j=1}^{n-1} y_{2j}}}+
underbrace{2sum_{j=1}^{n-1}y_{2j}color{red}{-4sum_{j=1}^{n-1} y_{2j}}}\
&=;;;;;;;;4sum_{j=1}^{2n-1}y_j
;;;;;;;;;;;;;;;;-2sum_{j=1}^{n-1}y_{2j}
;;;;;;=text{RHS}
end{align}$$
answered Jan 21 at 13:15
hypergeometrichypergeometric
17.7k1761
$endgroup$
$begingroup$
How do you add the first two sums in the second sentence of your formal proof?
$endgroup$
– GambitSquared
Jan 21 at 13:29
$begingroup$
The first is the sum of $y_{1,3,5,7,cdots,2n-1}$, and the second is the sum of $y_{2,4,6,cdots 2n-2}$. Adding them gives the sum of $y_{1,2,3,4,cdots,2n-1}$. All multiplied by $4$.
$endgroup$
– hypergeometric
Jan 21 at 13:37
add a comment |
$begingroup$
We obtain
begin{align*}
color{blue}{4sum_{j=1}^n}&color{blue}{y_{2j-1}+2sum_{j=1}^{n-1}y_{2j}}\
&=4left(sum_{j=1}^{2n-1}y_j-sum_{j=1}^{n-1}y_{2j}right)+2sum_{j=1}^{n-1}y_{2j}\
&,,color{blue}{=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}}\
end{align*}
and the claim follows.
answered Jan 22 at 20:38
Markus ScheuerMarkus Scheuer
62.2k459149
$endgroup$
add a comment |
$begingroup$
Note that
begin{align*}
4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}
&= 4(y_1+y_3+cdots+y_{2n-1})+2sum_{j=1}^{n-1}y_{2j} \
&= 4(y_1+y_2+cdots + y_{2n-2}+y_{2n-1}) - 4(y_2+y_4+cdots + y_{2n-2})+2sum_{j=1}^{n-1}y_{2j}\
&= 4sum_{j=1}^{2n-1}y_j - 4sum_{j=1}^{n-1} y_{2j}+2sum_{j=1}^{n-1}y_{2j}\
&= 4sum_{j=1}^{2n-1}y_j - 2sum_{j=1}^{n-1}y_{2j}.
end{align*}
answered Jan 20 at 16:01
rogerlrogerl
18k22747
$endgroup$
$begingroup$
Is it possible to manipulate the sums directly to show that the equality is true without expanding the sum?
$endgroup$
– GambitSquared
Jan 21 at 0:08
$begingroup$
Sure. I just expanded the sums to make it clear what was going on. You could equally well write it out by using sigma notation for the sums that I expanded in the answer.
$endgroup$
– rogerl
Jan 21 at 14:07
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%2f3080749%2fhow-to-show-that-4-sum-j-1ny-2j-12-sum-j-1n-1y-2j-equal-to-4-sum%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Maybe the multiple use of $j$ as index of summation is a bit confusing. Just note that
$$sum_{k=1}^{2n-1} y_k = underbrace{sum_{stackrel{k=1}{k: odd}}^{2n-1} y_k}_{=sum_{j=1}^{n} y_{2j-1} } + underbrace{sum_{stackrel{k=1}{k: even}}^{2n-1} y_k}_{=sum_{j=1}^{n-1} y_{2j}}$$
Now, you get immediately
$$4sum_{k=1}^{2n-1} y_k = 4 sum_{j=1}^{n} y_{2j-1} + underbrace{4}_{=2+2}sum_{j=1}^{n-1} y_{2j}Leftrightarrow$$
$$4sum_{k=1}^{2n-1} y_k - 2sum_{j=1}^{n-1} y_{2j} = 4 sum_{j=1}^{n} y_{2j-1} + 2sum_{j=1}^{n-1} y_{2j}$$
If you like you may now replace index $k$ by $j$ in the left-most sum.
answered Jan 21 at 11:50
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
Maybe the multiple use of $j$ as index of summation is a bit confusing. Just note that
$$sum_{k=1}^{2n-1} y_k = underbrace{sum_{stackrel{k=1}{k: odd}}^{2n-1} y_k}_{=sum_{j=1}^{n} y_{2j-1} } + underbrace{sum_{stackrel{k=1}{k: even}}^{2n-1} y_k}_{=sum_{j=1}^{n-1} y_{2j}}$$
Now, you get immediately
$$4sum_{k=1}^{2n-1} y_k = 4 sum_{j=1}^{n} y_{2j-1} + underbrace{4}_{=2+2}sum_{j=1}^{n-1} y_{2j}Leftrightarrow$$
$$4sum_{k=1}^{2n-1} y_k - 2sum_{j=1}^{n-1} y_{2j} = 4 sum_{j=1}^{n} y_{2j-1} + 2sum_{j=1}^{n-1} y_{2j}$$
If you like you may now replace index $k$ by $j$ in the left-most sum.
answered Jan 21 at 11:50
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
Maybe the multiple use of $j$ as index of summation is a bit confusing. Just note that
$$sum_{k=1}^{2n-1} y_k = underbrace{sum_{stackrel{k=1}{k: odd}}^{2n-1} y_k}_{=sum_{j=1}^{n} y_{2j-1} } + underbrace{sum_{stackrel{k=1}{k: even}}^{2n-1} y_k}_{=sum_{j=1}^{n-1} y_{2j}}$$
Now, you get immediately
$$4sum_{k=1}^{2n-1} y_k = 4 sum_{j=1}^{n} y_{2j-1} + underbrace{4}_{=2+2}sum_{j=1}^{n-1} y_{2j}Leftrightarrow$$
$$4sum_{k=1}^{2n-1} y_k - 2sum_{j=1}^{n-1} y_{2j} = 4 sum_{j=1}^{n} y_{2j-1} + 2sum_{j=1}^{n-1} y_{2j}$$
If you like you may now replace index $k$ by $j$ in the left-most sum.
answered Jan 21 at 11:50
trancelocationtrancelocation
12.6k1826
$endgroup$
Maybe the multiple use of $j$ as index of summation is a bit confusing. Just note that
$$sum_{k=1}^{2n-1} y_k = underbrace{sum_{stackrel{k=1}{k: odd}}^{2n-1} y_k}_{=sum_{j=1}^{n} y_{2j-1} } + underbrace{sum_{stackrel{k=1}{k: even}}^{2n-1} y_k}_{=sum_{j=1}^{n-1} y_{2j}}$$
Now, you get immediately
$$4sum_{k=1}^{2n-1} y_k = 4 sum_{j=1}^{n} y_{2j-1} + underbrace{4}_{=2+2}sum_{j=1}^{n-1} y_{2j}Leftrightarrow$$
$$4sum_{k=1}^{2n-1} y_k - 2sum_{j=1}^{n-1} y_{2j} = 4 sum_{j=1}^{n} y_{2j-1} + 2sum_{j=1}^{n-1} y_{2j}$$
If you like you may now replace index $k$ by $j$ in the left-most sum.
answered Jan 21 at 11:50
trancelocationtrancelocation
12.6k1826
answered Jan 21 at 11:50
trancelocationtrancelocation
12.6k1826
answered Jan 21 at 11:50
trancelocationtrancelocation
12.6k1826
answered Jan 21 at 11:50
trancelocationtrancelocation
12.6k1826
12.6k1826
add a comment |
add a comment |
$begingroup$
(NB - this is for illustration only)
$$4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}$$
As an illustration, take $n=4$.
LHS = $$begin{align}
4big(&y_1&&+y_3&&+y_5&&+y_7big)\
+2big(&&y_2&&+y_4&&+y_6&;;;;;;big)\
=;;;&4y_1&+2y_2&+4y_3&+2y_4&+4y_5&+2y_6&+4y_7end{align}$$
RHS=
$$begin{align}
4big(&y_1&+y_2&+y_3&+y_4&+y_5&y_6&+y_7big)\
-2big(&&y_2&&+y_4&&+y_6&;;;;;;big)\
=;;;&4y_1&+2y_2&+4y_3&+2y_4&+4y_5&+2y_6&+4y_7end{align}$$
Hence LHS=RHS.
Formal Proof
$$begin{align}
text{LHS}&=4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}\
&=underbrace{4sum_{j=1}^ny_{2j-1}color{red}{+4sum_{j=1}^{n-1} y_{2j}}}+
underbrace{2sum_{j=1}^{n-1}y_{2j}color{red}{-4sum_{j=1}^{n-1} y_{2j}}}\
&=;;;;;;;;4sum_{j=1}^{2n-1}y_j
;;;;;;;;;;;;;;;;-2sum_{j=1}^{n-1}y_{2j}
;;;;;;=text{RHS}
end{align}$$
answered Jan 21 at 13:15
hypergeometrichypergeometric
17.7k1761
$endgroup$
$begingroup$
How do you add the first two sums in the second sentence of your formal proof?
$endgroup$
– GambitSquared
Jan 21 at 13:29
$begingroup$
The first is the sum of $y_{1,3,5,7,cdots,2n-1}$, and the second is the sum of $y_{2,4,6,cdots 2n-2}$. Adding them gives the sum of $y_{1,2,3,4,cdots,2n-1}$. All multiplied by $4$.
$endgroup$
– hypergeometric
Jan 21 at 13:37
add a comment |
$begingroup$
(NB - this is for illustration only)
$$4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}$$
As an illustration, take $n=4$.
LHS = $$begin{align}
4big(&y_1&&+y_3&&+y_5&&+y_7big)\
+2big(&&y_2&&+y_4&&+y_6&;;;;;;big)\
=;;;&4y_1&+2y_2&+4y_3&+2y_4&+4y_5&+2y_6&+4y_7end{align}$$
RHS=
$$begin{align}
4big(&y_1&+y_2&+y_3&+y_4&+y_5&y_6&+y_7big)\
-2big(&&y_2&&+y_4&&+y_6&;;;;;;big)\
=;;;&4y_1&+2y_2&+4y_3&+2y_4&+4y_5&+2y_6&+4y_7end{align}$$
Hence LHS=RHS.
Formal Proof
$$begin{align}
text{LHS}&=4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}\
&=underbrace{4sum_{j=1}^ny_{2j-1}color{red}{+4sum_{j=1}^{n-1} y_{2j}}}+
underbrace{2sum_{j=1}^{n-1}y_{2j}color{red}{-4sum_{j=1}^{n-1} y_{2j}}}\
&=;;;;;;;;4sum_{j=1}^{2n-1}y_j
;;;;;;;;;;;;;;;;-2sum_{j=1}^{n-1}y_{2j}
;;;;;;=text{RHS}
end{align}$$
answered Jan 21 at 13:15
hypergeometrichypergeometric
17.7k1761
$endgroup$
$begingroup$
How do you add the first two sums in the second sentence of your formal proof?
$endgroup$
– GambitSquared
Jan 21 at 13:29
$begingroup$
The first is the sum of $y_{1,3,5,7,cdots,2n-1}$, and the second is the sum of $y_{2,4,6,cdots 2n-2}$. Adding them gives the sum of $y_{1,2,3,4,cdots,2n-1}$. All multiplied by $4$.
$endgroup$
– hypergeometric
Jan 21 at 13:37
add a comment |
$begingroup$
(NB - this is for illustration only)
$$4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}$$
As an illustration, take $n=4$.
LHS = $$begin{align}
4big(&y_1&&+y_3&&+y_5&&+y_7big)\
+2big(&&y_2&&+y_4&&+y_6&;;;;;;big)\
=;;;&4y_1&+2y_2&+4y_3&+2y_4&+4y_5&+2y_6&+4y_7end{align}$$
RHS=
$$begin{align}
4big(&y_1&+y_2&+y_3&+y_4&+y_5&y_6&+y_7big)\
-2big(&&y_2&&+y_4&&+y_6&;;;;;;big)\
=;;;&4y_1&+2y_2&+4y_3&+2y_4&+4y_5&+2y_6&+4y_7end{align}$$
Hence LHS=RHS.
Formal Proof
$$begin{align}
text{LHS}&=4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}\
&=underbrace{4sum_{j=1}^ny_{2j-1}color{red}{+4sum_{j=1}^{n-1} y_{2j}}}+
underbrace{2sum_{j=1}^{n-1}y_{2j}color{red}{-4sum_{j=1}^{n-1} y_{2j}}}\
&=;;;;;;;;4sum_{j=1}^{2n-1}y_j
;;;;;;;;;;;;;;;;-2sum_{j=1}^{n-1}y_{2j}
;;;;;;=text{RHS}
end{align}$$
answered Jan 21 at 13:15
hypergeometrichypergeometric
17.7k1761
$endgroup$
(NB - this is for illustration only)
$$4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}$$
As an illustration, take $n=4$.
LHS = $$begin{align}
4big(&y_1&&+y_3&&+y_5&&+y_7big)\
+2big(&&y_2&&+y_4&&+y_6&;;;;;;big)\
=;;;&4y_1&+2y_2&+4y_3&+2y_4&+4y_5&+2y_6&+4y_7end{align}$$
RHS=
$$begin{align}
4big(&y_1&+y_2&+y_3&+y_4&+y_5&y_6&+y_7big)\
-2big(&&y_2&&+y_4&&+y_6&;;;;;;big)\
=;;;&4y_1&+2y_2&+4y_3&+2y_4&+4y_5&+2y_6&+4y_7end{align}$$
Hence LHS=RHS.
Formal Proof
$$begin{align}
text{LHS}&=4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}\
&=underbrace{4sum_{j=1}^ny_{2j-1}color{red}{+4sum_{j=1}^{n-1} y_{2j}}}+
underbrace{2sum_{j=1}^{n-1}y_{2j}color{red}{-4sum_{j=1}^{n-1} y_{2j}}}\
&=;;;;;;;;4sum_{j=1}^{2n-1}y_j
;;;;;;;;;;;;;;;;-2sum_{j=1}^{n-1}y_{2j}
;;;;;;=text{RHS}
end{align}$$
answered Jan 21 at 13:15
hypergeometrichypergeometric
17.7k1761
edited Jan 21 at 13:27
answered Jan 21 at 13:15
hypergeometrichypergeometric
17.7k1761
answered Jan 21 at 13:15
hypergeometrichypergeometric
17.7k1761
answered Jan 21 at 13:15
hypergeometrichypergeometric
17.7k1761
17.7k1761
$begingroup$
How do you add the first two sums in the second sentence of your formal proof?
$endgroup$
– GambitSquared
Jan 21 at 13:29
$begingroup$
The first is the sum of $y_{1,3,5,7,cdots,2n-1}$, and the second is the sum of $y_{2,4,6,cdots 2n-2}$. Adding them gives the sum of $y_{1,2,3,4,cdots,2n-1}$. All multiplied by $4$.
$endgroup$
– hypergeometric
Jan 21 at 13:37
add a comment |
$begingroup$
How do you add the first two sums in the second sentence of your formal proof?
$endgroup$
– GambitSquared
Jan 21 at 13:29
$begingroup$
The first is the sum of $y_{1,3,5,7,cdots,2n-1}$, and the second is the sum of $y_{2,4,6,cdots 2n-2}$. Adding them gives the sum of $y_{1,2,3,4,cdots,2n-1}$. All multiplied by $4$.
$endgroup$
– hypergeometric
Jan 21 at 13:37
$begingroup$
How do you add the first two sums in the second sentence of your formal proof?
$endgroup$
– GambitSquared
Jan 21 at 13:29
$begingroup$
How do you add the first two sums in the second sentence of your formal proof?
$endgroup$
– GambitSquared
Jan 21 at 13:29
$begingroup$
The first is the sum of $y_{1,3,5,7,cdots,2n-1}$, and the second is the sum of $y_{2,4,6,cdots 2n-2}$. Adding them gives the sum of $y_{1,2,3,4,cdots,2n-1}$. All multiplied by $4$.
$endgroup$
– hypergeometric
Jan 21 at 13:37
$begingroup$
The first is the sum of $y_{1,3,5,7,cdots,2n-1}$, and the second is the sum of $y_{2,4,6,cdots 2n-2}$. Adding them gives the sum of $y_{1,2,3,4,cdots,2n-1}$. All multiplied by $4$.
$endgroup$
– hypergeometric
Jan 21 at 13:37
add a comment |
$begingroup$
We obtain
begin{align*}
color{blue}{4sum_{j=1}^n}&color{blue}{y_{2j-1}+2sum_{j=1}^{n-1}y_{2j}}\
&=4left(sum_{j=1}^{2n-1}y_j-sum_{j=1}^{n-1}y_{2j}right)+2sum_{j=1}^{n-1}y_{2j}\
&,,color{blue}{=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}}\
end{align*}
and the claim follows.
answered Jan 22 at 20:38
Markus ScheuerMarkus Scheuer
62.2k459149
$endgroup$
add a comment |
$begingroup$
We obtain
begin{align*}
color{blue}{4sum_{j=1}^n}&color{blue}{y_{2j-1}+2sum_{j=1}^{n-1}y_{2j}}\
&=4left(sum_{j=1}^{2n-1}y_j-sum_{j=1}^{n-1}y_{2j}right)+2sum_{j=1}^{n-1}y_{2j}\
&,,color{blue}{=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}}\
end{align*}
and the claim follows.
answered Jan 22 at 20:38
Markus ScheuerMarkus Scheuer
62.2k459149
$endgroup$
add a comment |
$begingroup$
We obtain
begin{align*}
color{blue}{4sum_{j=1}^n}&color{blue}{y_{2j-1}+2sum_{j=1}^{n-1}y_{2j}}\
&=4left(sum_{j=1}^{2n-1}y_j-sum_{j=1}^{n-1}y_{2j}right)+2sum_{j=1}^{n-1}y_{2j}\
&,,color{blue}{=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}}\
end{align*}
and the claim follows.
answered Jan 22 at 20:38
Markus ScheuerMarkus Scheuer
62.2k459149
$endgroup$
We obtain
begin{align*}
color{blue}{4sum_{j=1}^n}&color{blue}{y_{2j-1}+2sum_{j=1}^{n-1}y_{2j}}\
&=4left(sum_{j=1}^{2n-1}y_j-sum_{j=1}^{n-1}y_{2j}right)+2sum_{j=1}^{n-1}y_{2j}\
&,,color{blue}{=4sum_{j=1}^{2n-1}y_{j}-2sum_{j=1}^{n-1}y_{2j}}\
end{align*}
and the claim follows.
answered Jan 22 at 20:38
Markus ScheuerMarkus Scheuer
62.2k459149
answered Jan 22 at 20:38
Markus ScheuerMarkus Scheuer
62.2k459149
answered Jan 22 at 20:38
Markus ScheuerMarkus Scheuer
62.2k459149
answered Jan 22 at 20:38
Markus ScheuerMarkus Scheuer
62.2k459149
62.2k459149
add a comment |
add a comment |
$begingroup$
Note that
begin{align*}
4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}
&= 4(y_1+y_3+cdots+y_{2n-1})+2sum_{j=1}^{n-1}y_{2j} \
&= 4(y_1+y_2+cdots + y_{2n-2}+y_{2n-1}) - 4(y_2+y_4+cdots + y_{2n-2})+2sum_{j=1}^{n-1}y_{2j}\
&= 4sum_{j=1}^{2n-1}y_j - 4sum_{j=1}^{n-1} y_{2j}+2sum_{j=1}^{n-1}y_{2j}\
&= 4sum_{j=1}^{2n-1}y_j - 2sum_{j=1}^{n-1}y_{2j}.
end{align*}
answered Jan 20 at 16:01
rogerlrogerl
18k22747
$endgroup$
$begingroup$
Is it possible to manipulate the sums directly to show that the equality is true without expanding the sum?
$endgroup$
– GambitSquared
Jan 21 at 0:08
$begingroup$
Sure. I just expanded the sums to make it clear what was going on. You could equally well write it out by using sigma notation for the sums that I expanded in the answer.
$endgroup$
– rogerl
Jan 21 at 14:07
add a comment |
$begingroup$
Note that
begin{align*}
4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}
&= 4(y_1+y_3+cdots+y_{2n-1})+2sum_{j=1}^{n-1}y_{2j} \
&= 4(y_1+y_2+cdots + y_{2n-2}+y_{2n-1}) - 4(y_2+y_4+cdots + y_{2n-2})+2sum_{j=1}^{n-1}y_{2j}\
&= 4sum_{j=1}^{2n-1}y_j - 4sum_{j=1}^{n-1} y_{2j}+2sum_{j=1}^{n-1}y_{2j}\
&= 4sum_{j=1}^{2n-1}y_j - 2sum_{j=1}^{n-1}y_{2j}.
end{align*}
answered Jan 20 at 16:01
rogerlrogerl
18k22747
$endgroup$
$begingroup$
Is it possible to manipulate the sums directly to show that the equality is true without expanding the sum?
$endgroup$
– GambitSquared
Jan 21 at 0:08
$begingroup$
Sure. I just expanded the sums to make it clear what was going on. You could equally well write it out by using sigma notation for the sums that I expanded in the answer.
$endgroup$
– rogerl
Jan 21 at 14:07
add a comment |
$begingroup$
Note that
begin{align*}
4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}
&= 4(y_1+y_3+cdots+y_{2n-1})+2sum_{j=1}^{n-1}y_{2j} \
&= 4(y_1+y_2+cdots + y_{2n-2}+y_{2n-1}) - 4(y_2+y_4+cdots + y_{2n-2})+2sum_{j=1}^{n-1}y_{2j}\
&= 4sum_{j=1}^{2n-1}y_j - 4sum_{j=1}^{n-1} y_{2j}+2sum_{j=1}^{n-1}y_{2j}\
&= 4sum_{j=1}^{2n-1}y_j - 2sum_{j=1}^{n-1}y_{2j}.
end{align*}
answered Jan 20 at 16:01
rogerlrogerl
18k22747
$endgroup$
Note that
begin{align*}
4sum_{j=1}^ny_{2j-1}+2sum_{j=1}^{n-1}y_{2j}
&= 4(y_1+y_3+cdots+y_{2n-1})+2sum_{j=1}^{n-1}y_{2j} \
&= 4(y_1+y_2+cdots + y_{2n-2}+y_{2n-1}) - 4(y_2+y_4+cdots + y_{2n-2})+2sum_{j=1}^{n-1}y_{2j}\
&= 4sum_{j=1}^{2n-1}y_j - 4sum_{j=1}^{n-1} y_{2j}+2sum_{j=1}^{n-1}y_{2j}\
&= 4sum_{j=1}^{2n-1}y_j - 2sum_{j=1}^{n-1}y_{2j}.
end{align*}
answered Jan 20 at 16:01
rogerlrogerl
18k22747
edited Jan 20 at 20:42
answered Jan 20 at 16:01
rogerlrogerl
18k22747
answered Jan 20 at 16:01
rogerlrogerl
18k22747
answered Jan 20 at 16:01
rogerlrogerl
18k22747
18k22747
$begingroup$
Is it possible to manipulate the sums directly to show that the equality is true without expanding the sum?
$endgroup$
– GambitSquared
Jan 21 at 0:08
$begingroup$
Sure. I just expanded the sums to make it clear what was going on. You could equally well write it out by using sigma notation for the sums that I expanded in the answer.
$endgroup$
– rogerl
Jan 21 at 14:07
add a comment |
$begingroup$
Is it possible to manipulate the sums directly to show that the equality is true without expanding the sum?
$endgroup$
– GambitSquared
Jan 21 at 0:08
$begingroup$
Sure. I just expanded the sums to make it clear what was going on. You could equally well write it out by using sigma notation for the sums that I expanded in the answer.
$endgroup$
– rogerl
Jan 21 at 14:07
$begingroup$
Is it possible to manipulate the sums directly to show that the equality is true without expanding the sum?
$endgroup$
– GambitSquared
Jan 21 at 0:08
$begingroup$
Is it possible to manipulate the sums directly to show that the equality is true without expanding the sum?
$endgroup$
– GambitSquared
Jan 21 at 0:08
$begingroup$
Sure. I just expanded the sums to make it clear what was going on. You could equally well write it out by using sigma notation for the sums that I expanded in the answer.
$endgroup$
– rogerl
Jan 21 at 14:07
$begingroup$
Sure. I just expanded the sums to make it clear what was going on. You could equally well write it out by using sigma notation for the sums that I expanded in the answer.
$endgroup$
– rogerl
Jan 21 at 14:07
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%2f3080749%2fhow-to-show-that-4-sum-j-1ny-2j-12-sum-j-1n-1y-2j-equal-to-4-sum%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
