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 | |||||||||||
\(\ds \) | \(=\) | \(\ds n a_n + \sum_{k \mathop = 1}^{n - 1} k \left({a_k - a_{k + 1} }\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n a_n - \sum_{k \mathop = 1}^{n - 1} k \left({a_{k + 1} - a_k}\right)\) |
$\blacksquare$