Indexed Summation over Adjacent Intervals

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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.



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.

$\blacksquare$


Also see