Sum of Sequence as Summation of Difference of Adjacent Terms/Proof 2
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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
From Abel's Lemma: Formulation 2, after renaming and reassigning variables:
- $\ds \sum_{k \mathop = 1}^n a_k b_k = \sum_{k \mathop = 1}^{n - 1} \map {A_k} {a_k - a_{k + 1} } + A_n a_n$
where:
- $\sequence a$ and $\sequence b$ are sequences in $\R$
- $\ds A_n = \sum_{k \mathop = 1}^n {b_k}$ be the partial sum of $\sequence b$ from $1$ to $n$.
Let $\sequence b$ be defined as:
- $\forall k: b_k = 1$
Thus:
- $\ds A_n = \sum_{k \mathop = 1}^n 1 = n$
and so:
\(\ds \sum_{k \mathop = 1}^n a_k\) | \(=\) | \(\ds \sum_{k \mathop = 1}^{n - 1} k \paren {a_k - a_{k + 1} } + n a_n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n a_n - \sum_{k \mathop = 1}^{n - 1} k \paren {a_{k + 1} - a_k}\) |
$\blacksquare$