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?

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?

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TopNewestNo 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. – Mihail Tarigradschi · 4 years, 5 months ago

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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 · 4 years, 5 months ago

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aha, I have better understanding now. Thank you sire. – Shubham Raj · 4 years, 5 months ago

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