Sum of Sequence as Summation of Difference of Adjacent Terms

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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 1

\(\ds \sum_{k \mathop = 1}^n a_k\) \(=\) \(\ds \sum_{k \mathop = 1}^n \left({k - \left({k - 1}\right)}\right) a_k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n k a_k - \sum_{k \mathop = 1}^n \left({k - 1}\right) a_k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n k a_k - \sum_{k \mathop = 0}^{n - 1} k a_{k + 1}\) 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
\(\ds \) \(=\) \(\ds n a_n + \sum_{k \mathop = 1}^{n - 1} k \left({a_k - a_{k + 1} }\right)\)
\(\ds \) \(=\) \(\ds n a_n - \sum_{k \mathop = 1}^{n - 1} k \left({a_{k + 1} - a_k}\right)\)

$\blacksquare$


Proof 2

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:

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

$\blacksquare$


Sources

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