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# 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?

Note by Shubham Raj
4 years, 10 months ago

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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.

- 4 years, 10 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.

- 4 years, 10 months ago

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

- 4 years, 10 months ago