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