Sum of Sequence as Summation of Difference of Adjacent Terms/Proof 1

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Theorem

Let $n \in \Z_{> 0}$ be a strictly positive integer.

Then:

$\ds \sum_{k \mathop = 1}^n a_k = n a_n - \sum_{k \mathop = 1}^{n - 1} k \paren {a_{k + 1} - a_k}$


Proof

\(\ds \sum_{k \mathop = 1}^n a_k\) \(=\) \(\ds \sum_{k \mathop = 1}^n \left({k - \left({k - 1}\right)}\right) a_k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n k a_k - \sum_{k \mathop = 1}^n \left({k - 1}\right) a_k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n k a_k - \sum_{k \mathop = 0}^{n - 1} k a_{k + 1}\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds n a_n + \sum_{k \mathop = 1}^{n - 1} k a_k - \sum_{k \mathop = 1}^{n - 1} k a_{k + 1} + 0 \times a_0\) extracting the end terms
\(\ds \) \(=\) \(\ds n a_n + \sum_{k \mathop = 1}^{n - 1} k \left({a_k - a_{k + 1} }\right)\)
\(\ds \) \(=\) \(\ds n a_n - \sum_{k \mathop = 1}^{n - 1} k \left({a_{k + 1} - a_k}\right)\)

$\blacksquare$