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Formula for the sum of the first n integers

Prove that

    \[ \sum_{k=1}^n k = \frac{n^2}{2} + \frac{n}{2}. \]


Proof. From here we know that

    \[ \sum_{k=1}^n 1 = n \]

and from here, that

    \[ \sum_{k=1}^n (2k-1) = n^2. \]

By additivity and homogeneity of finite sums we have,

    \[ \sum_{k=1}^n (2k-1) = 2 \sum_{k=1}^n k - \sum_{k=1}^n 1 = n^2 \quad \implies \quad \sum_{k=1}^n k = \frac{n^2}{2} + \frac{n}{2}. \qquad \blacksquare \]

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