# A New Proof Of 1+2+3+...+n = $n(n+1)/2$

Let $f(n)=1+2+3+...+n$.

Now, we have $f(n)-f(n-1)=n$, and we will prove $f(n)+f(n-1)=n^2$ by induction.

Since the proposition is true for $n=1$, let us assume it is true for $n=k$. Now for $n=k+1$, we have: $f(k)+f(k+1)=[f(k)+f(k-1)]+2k+1=k^2+2k+1=(k+1)^2$. So, we conclude that $f(n)+f(n-1)=n^2$ for n belongs to N. Now, we have a system of simultaneous equations: $f(n)+f(n-1)=n^2$ $f(n)-f(n-1)=n$ solving them, we have $2f(n)=n^2+n$ which yields: $f(n)= \boxed {n(n+1)/2}$

Note by Mohammed Imran
1 year, 4 months ago

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I guess you meant f(n)-f(n-1)=n (not 2) at two instances.

- 1 year, 4 months ago

thank you very much!!!

- 1 year, 4 months ago

Cool

- 1 year, 3 months ago

thank you very much

- 1 year, 3 months ago

How is this proof?

- 1 year, 3 months ago

Well I guess that you used the mathematical induction in this problem, right? It's a kind of simple method to prove such these kind of mathematical statements. I like it.

- 1 year, 3 months ago

Thx!!!

- 1 year, 3 months ago

Cool proof! @Mohammed Imran

- 1 year, 2 months ago

thank you!

- 8 months, 3 weeks ago

Oh nice, where did you see this?

- 1 year, 4 months ago

this is the original proof of the theorem by gauss

- 1 year, 4 months ago

Then why did you type it as yours?🤔

- 1 year, 4 months ago

I said chew seong's proof is not original

- 1 year, 4 months ago

Ok

- 1 year, 4 months ago

@Brilliant Mathematics, @joan davis has wrote 5 posts that don't relate to mathematics.

- 1 year, 1 month ago

Thank you for notifying us again. We will take action on this shortly.

Staff - 1 year, 1 month ago