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Algebra (all content)
Course: Algebra (all content)ย >ย Unit 18
Lesson 2: Arithmetic series- Arithmetic series intro
- Arithmetic series
- Worked example: arithmetic series (sigma notation)
- Worked example: arithmetic series (sum expression)
- Worked example: arithmetic series (recursive formula)
- Arithmetic series worksheet
- Arithmetic series
- Proof of finite arithmetic series formula
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Proof of finite arithmetic series formula
Watch Sal prove the expression for the sum of all positive integers up to and including n. Created by Sal Khan.
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First off I find this kind of algebra awe inspiring. But I'm wondering where does the intuition for this problem come from? How did people originally figure out how to do this?(41 votes)
The formula for this probably dates back some one or two thousand years, but perhaps you will be interested by this amusing story:
It is said that Carl Friedrich Gauss, one of the greatest mathematicians of all time, was punished by a very mean teacher. His teacher wanted him to add up all the numbers from 1 to 100, thinking that it would take Gauss all afternoon. Of course, Gauss noticed that if he added 1 to 100, and 2 to 99, and 3 to 98, all the sums added up to 101. So, since you had 100 numbers, that means you had 50 pairs of numbers, that all added up to 101. The total sum is then a very easily computable 50 * 101 = 5050. There are many versions of this story, and it is not very clear if it is even true...
Like many mathematical results, it is often unclear from where they originate. (Many results in physics and calculus were discovered long before Newton published them) However, there are some amusing anecdotes like that Gauss story for many other results. For example, it is said that Napier invented the base of the natural logarithm because he wanted to know how much money he would make if it were compounded continuously.
I hope that although I couldn't answer your question, that I at least entertained you a bit with these stories.
Cheers(141 votes)
Well for example if I had a repeating dec mail what would I do(12 votes)
A repeating decimal is not an integer and would therefore not work with this rule.(26 votes)
Where did Sal prove it by induction?(10 votes)
https://www.khanacademy.org/math/algebra-home/precalculus/seq-induction/proof-by-induction/v/proof-by-induction
This video is a bit out of order, since the prove by induction is at the end of this topic.(11 votes)
I don't really understand how you came up with the second method.Where the s(n)=1+2+3....n+(n-1) is from? From the question or elsewhere.(4 votes)
S(N) = 1 + 2 + ...+(n-1) + n ; comes from the definition of the sum of n integers. It is defined to be the summation of your chosen integer and all preceding integers (ending at 1).
S(N) = n + (n-1) + ...+ 2 + 1; is the first equation written backwards, the reason for this is it becomes easier to see the pattern.
2(S(N)) = (n+1)n occurs when you add the corresponding pieces of the first and second S(N). It literally means that "To get double the summation" = "add the number n +1, n times"
Which is equivalent to S(N) = [(N+1)N]/2
For example: S(3) = n-2 + n-1 + n ==> 1 + 2 + 3 ==> ((3+1)3)1/2==>6
S(100) = n-99 + n-98 +....+n-1 + n ==> 1+2+...+99 +100 ==> ((100+1)100)1/2 ==>5050(13 votes)
Can you prove that 1ยฒ + 2ยฒ + 3ยฒ + ... + nยฒ = n(n+1)(2n+1)/6 ? Please Sal(5 votes)
step1.LHS=1 SO we subtitute n=1 INTO n(n+1)(2n+1)/6
LHS=RHS
STEP 2 assume that the formular is true for n=k
k=n(n+1)(5 votes)
If there are any videos related to this question- 1+3+5+.....+(2n-1 )= n^2
(principal of mathematical induction)
please give a link to the topic.(3 votes)
I think the only video regarding that topic is this one. You can use the same method shown in the video to prove your equality:S(n) = 1 + 3 + 5 + โฏ + (2n-5) + (2n-3) + (2n-1)
S(n) = (2n-1) + (2n-3) + (2n-5) + โฏ + 5 + 3 + 1
2S(n) = 2n + 2n + 2n + โฏ + 2n + 2n + 2n
2S(n) = (2n)ยทn
2S(n) = 2nยฒ
S(n) = nยฒ(7 votes)
can anybody please help me i need to know what is an induction?(3 votes)- What about S(nยฒ) including nยฒ ?(4 votes)
- Because nยฒ is an integer, and this proof applies to all integers, it would work for that one too. :)(1 vote)
- I was actually curious about this same question a few years ago and tried to see if i could come up with this formula. I did but i did it geometrically. So I imagined a square with some length n.and cut it up into n rows and n columns. if you represented each square as a number, you could represent the sum that sal is talking about by coloring in one square in the first column 2 squares in the second column 3 squares in the third column n squares in the nth column etc. then,since the diagonal of the square would be n squares, you could see that the sum is N^2/2 +n/2=N^2+n/2. its a lot easier to see it visually.(3 votes)
- This works only if the number of terms is same as the final number (1 to 100 in 100 terms). In fact, these two "n" are not same, should be written as "n" and "m".(2 votes)
- @1:10how 1+2+3+.........+(n-1)+n changed into n+(n-1)+(n-2)+........2+1(2 votes)
1:10Video transcript
In the last video, we
proved that the sum of all of the positive
integers up to and including n can be expressed as n
times n plus 1 over 2. And we proved that by induction. What I want to do in
this video is show you that there's actually a
simpler proof for that. But it's not by
induction, so it wouldn't be included in that video. But I'll show you
that it exists, just so you know that induction
isn't the only way to prove it. So we define that
function S of n as the sum of all of the
positive integers up to and including n. So this is equal
to, by definition, 1 plus 2 plus 3 plus, all the
way to plus n minus 1 plus n. So it's the sum of all of the
integers up to and including n. This is how we're defining it. Well, we can rewrite it again. We can say that the sum,
S of n-- we could just rewrite this same thing,
but we could rewrite it in a different order. We could say that this is the
same thing as n plus n minus 1 plus n minus 2 plus, all the
way down to plus 2 plus 1. Now, what does this do for us? Well, we can actually
add these two rows. If we add S of n
plus S of n, we're going to get 2 times
this sum, so we're just adding on the left. And then we can also
add on the right. So we're just adding this sum
twice, but what's interesting is how we're going to add it. We're going to add this
term to this term, this term to this term, because
we're really just trying to add these two things. And we can pick any way
we want to add them. So 1 plus n is going
to be n plus 1. And then we're going to add--
let me do it in pink-- 2 plus n minus 1. So what's 2 plus n minus 1? Let me write it over here. 2 plus n minus 1. It's the same thing
as 2 plus n minus 1, which is the same
thing as n plus 1. 2 minus 1 is just 1. So this is also
going to be n plus 1. And then this term over here,
3 plus n minus 2, or n minus 2 plus 3. Once again, that's
going to be n plus 1. And you're going to
do that for every term all the way until you get
over here, n minus 1 plus 2. That's also going
to be n plus 1. And then finally, you have
n plus 1 right over here. Plus n plus 1. So what's this whole
sum going to be? Well, how many of these
n plus 1's do we have? Well, we have n of
them, for every term in each of these sums. So this is 1, 2, 3,
count all the way to n. You have n of these terms. So you have n n plus 1's. So if you add something
to itself n times, or if you have something
n times right over here, this is exactly equivalent
to n times n plus 1. So 2 times that sum of
all the positive integers up to and including
n is going to be equal to n times n plus 1. So if you divide
both sides by 2, we get an expression
for the sum. So the sum of all the positive
integers up to and including n is going to be equal to
n times n plus 1 over 2. So here was a proof where we
didn't have to use induction. It's really kind of a
pure algebraic proof.