Solution :
Let \(S_n\) = 1 + 2 + 3 + ….. + n = \(\sum_{k=1}^{n} k\)
Clearly, it is an arithmetic series with first term a = 1, common difference d = 1 and last term l = n.
\(\therefore\) \(S_n\) = \(n\over 2\) (1 + n) = \(n(n+1)\over 2\)
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