Indexed Summation without First Term
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Theorem
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $a$ and $b$ be integers with $a \le b$.
Let $\closedint a b$ be the integer interval between $a$ and $b$.
Let $f: \closedint a b \to \mathbb A$ be a mapping.
Then we have an equality of indexed summations:
- $\ds \sum_{i \mathop = a}^b \map f i = \map f a + \sum_{i \mathop = a + 1}^b \map f {\map \sigma i}$
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Proof
The proof goes by induction on $b$.
Basis for the Induction
Let $b = a$.
We have:
\(\ds \sum_{i \mathop = a}^a \map f i\) | \(=\) | \(\ds \map f a\) | Indexed Summation over Interval of Length One | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f a + 0\) | Identity Element of Addition on Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f a + \sum_{i \mathop = a + 1}^a \map f i\) | Definition of Indexed Summation, $a + 1 > a$ |
This is our basis for the induction.
Induction Step
Let $b \ge a + 1$.
We have:
\(\ds \sum_{i \mathop = a}^b \map f i\) | \(=\) | \(\ds \sum_{i \mathop = a}^{b - 1} \map f i + \map f b\) | Definition of Indexed Summation, $b \ge a$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f a + \sum_{i \mathop = a + 1}^{b - 1} \map f i + \map f b\) | induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f a + \sum_{i \mathop = a + 1}^b \map f i\) | Definition of Indexed Summation, $b \ge a + 1$ |
By the Principle of Mathematical Induction, the proof is complete.
$\blacksquare$