Combinatorics and sum of natural numbers

We all know that

C(n+1 ,2) = n(n+1)/2 and 1+2+3...n = n(n+1)/2

My question is, is it just a co-incidence that these 2 are equal ? Or can combinations be used to derive the formula for sum of first n natural numbers?

#Combinatorics #NumberTheory #Logic

Easy Math Editor

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Comments

No actually, if you understand the meaning behind $C(n+1,2)$ it is the number of distinct pairs between $n+1$ numbers, consider those numbers to be $1,2,...,n+1$ now you can pair $1$ with $2,3,...,n+1$ , counting them we get $n+1-1$ pairs that contain $1$ , doing the same thing for $2,...,n+1$ while ignoring equal pairs we get $C(n+1,2)=n+n-1+...+1=n(n+1)/2$ . Hope it helped.

A Former Brilliant Member - 8 years, 3 months ago

aha, I have better understanding now. Thank you sire.

Shubham Raj - 8 years, 3 months ago

yes we can do say suppose we have to sum first 10 natural numbers we can C(11,2)=11*10/2=55 in this way we can do alot of sums.

Niaz Ghumro - 8 years, 3 months ago

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}`	$\boxed{123}$