If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Algebra 2 (Eureka Math/EngageNY)>Unit 3

Lesson 13: Topic E: Lesson 29: Arithmetic series

# 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.

## Want to join the conversation?

• 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?
• 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
• Well for example if I had a repeating dec mail what would I do
• A repeating decimal is not an integer and would therefore not work with this rule.
• Where did Sal prove it by induction?
• 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.
• 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
• Can you prove that 1² + 2² + 3² + ... + n² = n(n+1)(2n+1)/6 ? Please Sal
• 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)
• If there are any videos related to this question- 1+3+5+.....+(2n-1 )= n^2
(principal of mathematical induction)
• 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 + ⋯ + (2n-5) + (2n-3) + (2n-1)S(n) = (2n-1) + (2n-3) + (2n-5) + ⋯ + 5 + 3 + 12S(n) = 2n + 2n + 2n + ⋯ + 2n + 2n + 2n2S(n) = (2n)·n2S(n) = 2n²S(n) = n²``
• What about S(n²) including n² ?