Steadily accumulating number formula
$begingroup$
This is quite likely a super simple question, but I lack the descriptive power to be able to find the answer on Google!
Consider the following two number sequences:
00, 10, 20, 30, 40, 50, 60, 70, 80, 90
45, 45, 45, 45, 45, 45, 45, 45, 45, 45
Both sequences have a total value of 450, but one accumulates steadily in multiples of 10, whilst the second remains constant at 45.
The second is easy to ascertain, divide the total value by the number of instances in the list. But the first sequence I'm stuck on - can anyone help me understand the formula for ascertaining the accumulation value (in this instance 10) from 0?
My goal is to be able to take any number and divide it x number of times, with the first number in the sequence being 0 and steadily increasing - apportioning greater and greater values to those later in the sequence.
sequences-and-series
asked Mar 25 at 4:24
Ryan GilliesRyan Gillies
182
New contributor
Ryan Gillies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This is quite likely a super simple question, but I lack the descriptive power to be able to find the answer on Google!
Consider the following two number sequences:
00, 10, 20, 30, 40, 50, 60, 70, 80, 90
45, 45, 45, 45, 45, 45, 45, 45, 45, 45
Both sequences have a total value of 450, but one accumulates steadily in multiples of 10, whilst the second remains constant at 45.
The second is easy to ascertain, divide the total value by the number of instances in the list. But the first sequence I'm stuck on - can anyone help me understand the formula for ascertaining the accumulation value (in this instance 10) from 0?
My goal is to be able to take any number and divide it x number of times, with the first number in the sequence being 0 and steadily increasing - apportioning greater and greater values to those later in the sequence.
sequences-and-series
asked Mar 25 at 4:24
Ryan GilliesRyan Gillies
182
New contributor
Ryan Gillies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
The keywords you need are "arithmetic progression".
$endgroup$
– FredH
Mar 25 at 4:33
$begingroup$
Thank you, that instantly helps a lot with my googling!
$endgroup$
– Ryan Gillies
Mar 25 at 4:39
add a comment |
$begingroup$
This is quite likely a super simple question, but I lack the descriptive power to be able to find the answer on Google!
Consider the following two number sequences:
00, 10, 20, 30, 40, 50, 60, 70, 80, 90
45, 45, 45, 45, 45, 45, 45, 45, 45, 45
Both sequences have a total value of 450, but one accumulates steadily in multiples of 10, whilst the second remains constant at 45.
The second is easy to ascertain, divide the total value by the number of instances in the list. But the first sequence I'm stuck on - can anyone help me understand the formula for ascertaining the accumulation value (in this instance 10) from 0?
My goal is to be able to take any number and divide it x number of times, with the first number in the sequence being 0 and steadily increasing - apportioning greater and greater values to those later in the sequence.
sequences-and-series
asked Mar 25 at 4:24
Ryan GilliesRyan Gillies
182
New contributor
Ryan Gillies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This is quite likely a super simple question, but I lack the descriptive power to be able to find the answer on Google!
Consider the following two number sequences:
00, 10, 20, 30, 40, 50, 60, 70, 80, 90
45, 45, 45, 45, 45, 45, 45, 45, 45, 45
Both sequences have a total value of 450, but one accumulates steadily in multiples of 10, whilst the second remains constant at 45.
The second is easy to ascertain, divide the total value by the number of instances in the list. But the first sequence I'm stuck on - can anyone help me understand the formula for ascertaining the accumulation value (in this instance 10) from 0?
My goal is to be able to take any number and divide it x number of times, with the first number in the sequence being 0 and steadily increasing - apportioning greater and greater values to those later in the sequence.
sequences-and-series
sequences-and-series
asked Mar 25 at 4:24
Ryan GilliesRyan Gillies
182
New contributor
Ryan Gillies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 25 at 4:24
Ryan GilliesRyan Gillies
182
New contributor
Ryan Gillies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 25 at 4:24
Ryan GilliesRyan Gillies
182
New contributor
Ryan Gillies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 25 at 4:24
Ryan GilliesRyan Gillies
182
asked Mar 25 at 4:24
Ryan GilliesRyan Gillies
182
182
New contributor
Ryan Gillies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ryan Gillies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ryan Gillies is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
The keywords you need are "arithmetic progression".
$endgroup$
– FredH
Mar 25 at 4:33
$begingroup$
Thank you, that instantly helps a lot with my googling!
$endgroup$
– Ryan Gillies
Mar 25 at 4:39
add a comment |
$begingroup$
The keywords you need are "arithmetic progression".
$endgroup$
– FredH
Mar 25 at 4:33
$begingroup$
Thank you, that instantly helps a lot with my googling!
$endgroup$
– Ryan Gillies
Mar 25 at 4:39
$begingroup$
The keywords you need are "arithmetic progression".
$endgroup$
– FredH
Mar 25 at 4:33
$begingroup$
The keywords you need are "arithmetic progression".
$endgroup$
– FredH
Mar 25 at 4:33
$begingroup$
Thank you, that instantly helps a lot with my googling!
$endgroup$
– Ryan Gillies
Mar 25 at 4:39
$begingroup$
Thank you, that instantly helps a lot with my googling!
$endgroup$
– Ryan Gillies
Mar 25 at 4:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's an arithmetic sequence (i. e. every next element is created from the previous with adding the same number) with the well-known formula for its sum
$$a_1 + a_2 + cdots + a_n = {n over 2}(a_1+a_n)$$
So, as $a_1= 0, n = 10, $ and $ a_{10} = 90$ you will obtain
$$00 + 10 +20 + cdots + 90 = {10over2}(0 + 90)= 5cdot 90 = 450$$
Addendum:
If you know the first element ($a_1$), number of elements ($n$) and their sum ($s$), you may obtain the last element ($a_n$) by substituting the known values into the formula, and then isolate $a_n$:
$$a_n = {2sover n}-a_1$$
You may then compute the difference $d$ between adjacent members as
$$d = {a_n-a_1over n-1}$$
to consecutively obtain members between $a_1$ and $a_n$:
begin{aligned}
a_2 &= a_1+d\a_3 &= a_2+d\
&cdots
end{aligned}
(To the question in your comment.)
answered Mar 25 at 4:37
MarianDMarianD
1,7791617
$endgroup$
$begingroup$
Using this approach it appears I need to know the answer toan(90) - if I only havea1(0) and 450, could it still be done?
$endgroup$
– Ryan Gillies
Mar 25 at 4:43
$begingroup$
@Ryan You have the values for n, $a_1$, and the sum. That leaves $a_n$ as the only variable. You can solve for $a_n$ to get $a_n=frac{2}{n}(x)-a_1$ where $x$ is the sum (450 in your case).
$endgroup$
– Jacob Jones
Mar 25 at 4:57
$begingroup$
Sorry yes you're right I forgot to include that in my comment, I will always know number of elements (n) first number in sequence (a1) and the sum (in this example 450). What I don't know is the last number in sequence (an).
$endgroup$
– Ryan Gillies
Mar 25 at 4:57
$begingroup$
You need know also the number of elements ($n$) to substitute it to the given formula, because without this you may write the sum $450$ in many different ways, e. g. $450 = 0 + 450 = 0 + 150 + 300 = 0 +45 +90 +135 +180$.
$endgroup$
– MarianD
Mar 25 at 5:00
$begingroup$
Nailed it, understood perfectly! Thank you so much for your help MarianD and @JacobJones!
$endgroup$
– Ryan Gillies
Mar 25 at 5:01
add a comment |
$begingroup$
$$sumlimits_{i=0}^9 10 i = 10 sumlimits_{i=0}^9 i = 10 left[ frac{(9+0)(9+1)}{2} right] =450$$
answered Mar 25 at 4:34
David G. StorkDavid G. Stork
11.5k41533
$endgroup$
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
});
}
});
Ryan Gillies is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3161382%2fsteadily-accumulating-number-formula%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
$begingroup$
It's an arithmetic sequence (i. e. every next element is created from the previous with adding the same number) with the well-known formula for its sum
$$a_1 + a_2 + cdots + a_n = {n over 2}(a_1+a_n)$$
So, as $a_1= 0, n = 10, $ and $ a_{10} = 90$ you will obtain
$$00 + 10 +20 + cdots + 90 = {10over2}(0 + 90)= 5cdot 90 = 450$$
Addendum:
If you know the first element ($a_1$), number of elements ($n$) and their sum ($s$), you may obtain the last element ($a_n$) by substituting the known values into the formula, and then isolate $a_n$:
$$a_n = {2sover n}-a_1$$
You may then compute the difference $d$ between adjacent members as
$$d = {a_n-a_1over n-1}$$
to consecutively obtain members between $a_1$ and $a_n$:
begin{aligned}
a_2 &= a_1+d\a_3 &= a_2+d\
&cdots
end{aligned}
(To the question in your comment.)
answered Mar 25 at 4:37
MarianDMarianD
1,7791617
$endgroup$
$begingroup$
Using this approach it appears I need to know the answer toan(90) - if I only havea1(0) and 450, could it still be done?
$endgroup$
– Ryan Gillies
Mar 25 at 4:43
$begingroup$
@Ryan You have the values for n, $a_1$, and the sum. That leaves $a_n$ as the only variable. You can solve for $a_n$ to get $a_n=frac{2}{n}(x)-a_1$ where $x$ is the sum (450 in your case).
$endgroup$
– Jacob Jones
Mar 25 at 4:57
$begingroup$
Sorry yes you're right I forgot to include that in my comment, I will always know number of elements (n) first number in sequence (a1) and the sum (in this example 450). What I don't know is the last number in sequence (an).
$endgroup$
– Ryan Gillies
Mar 25 at 4:57
$begingroup$
You need know also the number of elements ($n$) to substitute it to the given formula, because without this you may write the sum $450$ in many different ways, e. g. $450 = 0 + 450 = 0 + 150 + 300 = 0 +45 +90 +135 +180$.
$endgroup$
– MarianD
Mar 25 at 5:00
$begingroup$
Nailed it, understood perfectly! Thank you so much for your help MarianD and @JacobJones!
$endgroup$
– Ryan Gillies
Mar 25 at 5:01
add a comment |
$begingroup$
It's an arithmetic sequence (i. e. every next element is created from the previous with adding the same number) with the well-known formula for its sum
$$a_1 + a_2 + cdots + a_n = {n over 2}(a_1+a_n)$$
So, as $a_1= 0, n = 10, $ and $ a_{10} = 90$ you will obtain
$$00 + 10 +20 + cdots + 90 = {10over2}(0 + 90)= 5cdot 90 = 450$$
Addendum:
If you know the first element ($a_1$), number of elements ($n$) and their sum ($s$), you may obtain the last element ($a_n$) by substituting the known values into the formula, and then isolate $a_n$:
$$a_n = {2sover n}-a_1$$
You may then compute the difference $d$ between adjacent members as
$$d = {a_n-a_1over n-1}$$
to consecutively obtain members between $a_1$ and $a_n$:
begin{aligned}
a_2 &= a_1+d\a_3 &= a_2+d\
&cdots
end{aligned}
(To the question in your comment.)
answered Mar 25 at 4:37
MarianDMarianD
1,7791617
$endgroup$
$begingroup$
Using this approach it appears I need to know the answer toan(90) - if I only havea1(0) and 450, could it still be done?
$endgroup$
– Ryan Gillies
Mar 25 at 4:43
$begingroup$
@Ryan You have the values for n, $a_1$, and the sum. That leaves $a_n$ as the only variable. You can solve for $a_n$ to get $a_n=frac{2}{n}(x)-a_1$ where $x$ is the sum (450 in your case).
$endgroup$
– Jacob Jones
Mar 25 at 4:57
$begingroup$
Sorry yes you're right I forgot to include that in my comment, I will always know number of elements (n) first number in sequence (a1) and the sum (in this example 450). What I don't know is the last number in sequence (an).
$endgroup$
– Ryan Gillies
Mar 25 at 4:57
$begingroup$
You need know also the number of elements ($n$) to substitute it to the given formula, because without this you may write the sum $450$ in many different ways, e. g. $450 = 0 + 450 = 0 + 150 + 300 = 0 +45 +90 +135 +180$.
$endgroup$
– MarianD
Mar 25 at 5:00
$begingroup$
Nailed it, understood perfectly! Thank you so much for your help MarianD and @JacobJones!
$endgroup$
– Ryan Gillies
Mar 25 at 5:01
add a comment |
$begingroup$
It's an arithmetic sequence (i. e. every next element is created from the previous with adding the same number) with the well-known formula for its sum
$$a_1 + a_2 + cdots + a_n = {n over 2}(a_1+a_n)$$
So, as $a_1= 0, n = 10, $ and $ a_{10} = 90$ you will obtain
$$00 + 10 +20 + cdots + 90 = {10over2}(0 + 90)= 5cdot 90 = 450$$
Addendum:
If you know the first element ($a_1$), number of elements ($n$) and their sum ($s$), you may obtain the last element ($a_n$) by substituting the known values into the formula, and then isolate $a_n$:
$$a_n = {2sover n}-a_1$$
You may then compute the difference $d$ between adjacent members as
$$d = {a_n-a_1over n-1}$$
to consecutively obtain members between $a_1$ and $a_n$:
begin{aligned}
a_2 &= a_1+d\a_3 &= a_2+d\
&cdots
end{aligned}
(To the question in your comment.)
answered Mar 25 at 4:37
MarianDMarianD
1,7791617
$endgroup$
It's an arithmetic sequence (i. e. every next element is created from the previous with adding the same number) with the well-known formula for its sum
$$a_1 + a_2 + cdots + a_n = {n over 2}(a_1+a_n)$$
So, as $a_1= 0, n = 10, $ and $ a_{10} = 90$ you will obtain
$$00 + 10 +20 + cdots + 90 = {10over2}(0 + 90)= 5cdot 90 = 450$$
Addendum:
If you know the first element ($a_1$), number of elements ($n$) and their sum ($s$), you may obtain the last element ($a_n$) by substituting the known values into the formula, and then isolate $a_n$:
$$a_n = {2sover n}-a_1$$
You may then compute the difference $d$ between adjacent members as
$$d = {a_n-a_1over n-1}$$
to consecutively obtain members between $a_1$ and $a_n$:
begin{aligned}
a_2 &= a_1+d\a_3 &= a_2+d\
&cdots
end{aligned}
(To the question in your comment.)
answered Mar 25 at 4:37
MarianDMarianD
1,7791617
edited Mar 25 at 5:24
answered Mar 25 at 4:37
MarianDMarianD
1,7791617
answered Mar 25 at 4:37
MarianDMarianD
1,7791617
answered Mar 25 at 4:37
MarianDMarianD
1,7791617
1,7791617
$begingroup$
Using this approach it appears I need to know the answer toan(90) - if I only havea1(0) and 450, could it still be done?
$endgroup$
– Ryan Gillies
Mar 25 at 4:43
$begingroup$
@Ryan You have the values for n, $a_1$, and the sum. That leaves $a_n$ as the only variable. You can solve for $a_n$ to get $a_n=frac{2}{n}(x)-a_1$ where $x$ is the sum (450 in your case).
$endgroup$
– Jacob Jones
Mar 25 at 4:57
$begingroup$
Sorry yes you're right I forgot to include that in my comment, I will always know number of elements (n) first number in sequence (a1) and the sum (in this example 450). What I don't know is the last number in sequence (an).
$endgroup$
– Ryan Gillies
Mar 25 at 4:57
$begingroup$
You need know also the number of elements ($n$) to substitute it to the given formula, because without this you may write the sum $450$ in many different ways, e. g. $450 = 0 + 450 = 0 + 150 + 300 = 0 +45 +90 +135 +180$.
$endgroup$
– MarianD
Mar 25 at 5:00
$begingroup$
Nailed it, understood perfectly! Thank you so much for your help MarianD and @JacobJones!
$endgroup$
– Ryan Gillies
Mar 25 at 5:01
add a comment |
$begingroup$
Using this approach it appears I need to know the answer toan(90) - if I only havea1(0) and 450, could it still be done?
$endgroup$
– Ryan Gillies
Mar 25 at 4:43
$begingroup$
@Ryan You have the values for n, $a_1$, and the sum. That leaves $a_n$ as the only variable. You can solve for $a_n$ to get $a_n=frac{2}{n}(x)-a_1$ where $x$ is the sum (450 in your case).
$endgroup$
– Jacob Jones
Mar 25 at 4:57
$begingroup$
Sorry yes you're right I forgot to include that in my comment, I will always know number of elements (n) first number in sequence (a1) and the sum (in this example 450). What I don't know is the last number in sequence (an).
$endgroup$
– Ryan Gillies
Mar 25 at 4:57
$begingroup$
You need know also the number of elements ($n$) to substitute it to the given formula, because without this you may write the sum $450$ in many different ways, e. g. $450 = 0 + 450 = 0 + 150 + 300 = 0 +45 +90 +135 +180$.
$endgroup$
– MarianD
Mar 25 at 5:00
$begingroup$
Nailed it, understood perfectly! Thank you so much for your help MarianD and @JacobJones!
$endgroup$
– Ryan Gillies
Mar 25 at 5:01
$begingroup$
Using this approach it appears I need to know the answer to
an (90) - if I only have a1 (0) and 450, could it still be done?$endgroup$
– Ryan Gillies
Mar 25 at 4:43
$begingroup$
Using this approach it appears I need to know the answer to
an (90) - if I only have a1 (0) and 450, could it still be done?$endgroup$
– Ryan Gillies
Mar 25 at 4:43
$begingroup$
@Ryan You have the values for n, $a_1$, and the sum. That leaves $a_n$ as the only variable. You can solve for $a_n$ to get $a_n=frac{2}{n}(x)-a_1$ where $x$ is the sum (450 in your case).
$endgroup$
– Jacob Jones
Mar 25 at 4:57
$begingroup$
@Ryan You have the values for n, $a_1$, and the sum. That leaves $a_n$ as the only variable. You can solve for $a_n$ to get $a_n=frac{2}{n}(x)-a_1$ where $x$ is the sum (450 in your case).
$endgroup$
– Jacob Jones
Mar 25 at 4:57
$begingroup$
Sorry yes you're right I forgot to include that in my comment, I will always know number of elements (n) first number in sequence (a1) and the sum (in this example 450). What I don't know is the last number in sequence (an).
$endgroup$
– Ryan Gillies
Mar 25 at 4:57
$begingroup$
Sorry yes you're right I forgot to include that in my comment, I will always know number of elements (n) first number in sequence (a1) and the sum (in this example 450). What I don't know is the last number in sequence (an).
$endgroup$
– Ryan Gillies
Mar 25 at 4:57
$begingroup$
You need know also the number of elements ($n$) to substitute it to the given formula, because without this you may write the sum $450$ in many different ways, e. g. $450 = 0 + 450 = 0 + 150 + 300 = 0 +45 +90 +135 +180$.
$endgroup$
– MarianD
Mar 25 at 5:00
$begingroup$
You need know also the number of elements ($n$) to substitute it to the given formula, because without this you may write the sum $450$ in many different ways, e. g. $450 = 0 + 450 = 0 + 150 + 300 = 0 +45 +90 +135 +180$.
$endgroup$
– MarianD
Mar 25 at 5:00
$begingroup$
Nailed it, understood perfectly! Thank you so much for your help MarianD and @JacobJones!
$endgroup$
– Ryan Gillies
Mar 25 at 5:01
$begingroup$
Nailed it, understood perfectly! Thank you so much for your help MarianD and @JacobJones!
$endgroup$
– Ryan Gillies
Mar 25 at 5:01
add a comment |
$begingroup$
$$sumlimits_{i=0}^9 10 i = 10 sumlimits_{i=0}^9 i = 10 left[ frac{(9+0)(9+1)}{2} right] =450$$
answered Mar 25 at 4:34
David G. StorkDavid G. Stork
11.5k41533
$endgroup$
add a comment |
$begingroup$
$$sumlimits_{i=0}^9 10 i = 10 sumlimits_{i=0}^9 i = 10 left[ frac{(9+0)(9+1)}{2} right] =450$$
answered Mar 25 at 4:34
David G. StorkDavid G. Stork
11.5k41533
$endgroup$
add a comment |
$begingroup$
$$sumlimits_{i=0}^9 10 i = 10 sumlimits_{i=0}^9 i = 10 left[ frac{(9+0)(9+1)}{2} right] =450$$
answered Mar 25 at 4:34
David G. StorkDavid G. Stork
11.5k41533
$endgroup$
$$sumlimits_{i=0}^9 10 i = 10 sumlimits_{i=0}^9 i = 10 left[ frac{(9+0)(9+1)}{2} right] =450$$
answered Mar 25 at 4:34
David G. StorkDavid G. Stork
11.5k41533
answered Mar 25 at 4:34
David G. StorkDavid G. Stork
11.5k41533
answered Mar 25 at 4:34
David G. StorkDavid G. Stork
11.5k41533
answered Mar 25 at 4:34
David G. StorkDavid G. Stork
11.5k41533
11.5k41533
add a comment |
add a comment |
Ryan Gillies is a new contributor. Be nice, and check out our Code of Conduct.
Ryan Gillies is a new contributor. Be nice, and check out our Code of Conduct.
Ryan Gillies is a new contributor. Be nice, and check out our Code of Conduct.
Ryan Gillies is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3161382%2fsteadily-accumulating-number-formula%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
$begingroup$
The keywords you need are "arithmetic progression".
$endgroup$
– FredH
Mar 25 at 4:33
$begingroup$
Thank you, that instantly helps a lot with my googling!
$endgroup$
– Ryan Gillies
Mar 25 at 4:39