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: $f(n)= \boxed {n(n+1)/2}$

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

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

Comments

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