# Indexed Summation of Multiple of Mapping

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## Theorem

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.

Let $a, b$ be integers.

Let $\closedint a b$ denote the integer interval between $a$ and $b$.

Let $f: \closedint a b \to \mathbb A$ be a mapping.

Let $\lambda \in \mathbb A$.

Let $g = \lambda \cdot f$ be the product of $f$ with $\lambda$.

Then we have the equality of indexed summations:

- $\ds \sum_{i \mathop = a}^b \map g i = \lambda \cdot \sum_{i \mathop = a}^b \map f i$

## Proof

The proof goes by induction on $b$.

### Basis for the Induction

Let $b < a$.

Then all indexed summations are zero.

Because $0 = \lambda \cdot 0$, the result follows.

This is our basis for the induction.

### Induction Step

Let $b \geq a$.

We have:

\(\ds \sum_{i \mathop = a}^b \map g i\) | \(=\) | \(\ds \sum_{i \mathop = a}^{b - 1} \map g i + \map g b\) | Definition of Indexed Summation | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = a}^{b - 1} \map g i + \lambda \cdot \map f b\) | Definition of Product of Mapping with Scalar | |||||||||||

\(\ds \) | \(=\) | \(\ds \lambda \cdot \sum_{i \mathop = a}^{b - 1} \map f i + \lambda \cdot \map fb\) | Induction Hypothesis | |||||||||||

\(\ds \) | \(=\) | \(\ds \lambda \cdot \paren {\sum_{i \mathop = a}^{b - 1} \map f i + \map f b}\) | Distributive Property | |||||||||||

\(\ds \) | \(=\) | \(\ds \lambda \cdot \sum_{i \mathop = a}^b \map f i\) | Definition of Indexed Summation |

By the Principle of Mathematical Induction, the proof is complete.

$\blacksquare$