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Prove that the sum of 1’s from 1 to n equals n

Prove: \displaystyle{\sum_{k=1}^n 1 = n}.


Proof. We prove by induction. If n = 1, then the statement is true since \sum_{k=1}^1 1 = 1. Assume then that the statement is true for some n = m \in \mathbb{Z}_{>0}. So,

    \[  \sum_{k=1}^m 1 = m \quad \implies \quad \left( \sum_{k=1}^m 1 \right) + 1 = m+1 \quad \implies \quad \sum_{k=1}^{m+1} 1 = m+1. \]

Where the final step follows since the value of 1 for k = m +1 is still 1. (Maybe this could confuse since we are summing over the index k, but the value is independent of k. So, really, we are just counting… so for each k in the index we add 1; thus, when we have the sum from k = 1 to m and add 1, it is the same as summing from k = 1 to m+1). Thus, by induction, the statement is true for all n \in \mathbb{Z}_{> 0}. \qquad \blacksquare

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