# Summation is Linear/Sum of Summations

## Theorem

Let $\left({x_1, \ldots, x_n}\right)$ and $\left({y_1, \ldots, y_n}\right)$ be finite sequences of numbers of equal length.

Let $\lambda$ be a number.

Then:

- $\displaystyle \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \left({x_i + y_i}\right)$

## Proof

The proof proceeds by mathematical induction.

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:

- $\displaystyle \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \left({x_i + y_i}\right)$

### Basis for the Induction

$P \left({1}\right)$ is the case:

\(\displaystyle \sum_{i \mathop = 1}^1 x_i + \sum_{i \mathop = 1}^1 y_i\) | \(=\) | \(\displaystyle x_1 + y_1\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^1 \left({x_i + y_i}\right)\) |

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:

- $\displaystyle \sum_{i \mathop = 1}^k x_i + \sum_{i \mathop = 1}^k y_i = \sum_{i \mathop = 1}^k \left({x_i + y_i}\right)$

from which it is to be shown that:

- $\displaystyle \sum_{i \mathop = 1}^{k + 1} x_i + \sum_{i \mathop = 1}^{k + 1} y_i = \sum_{i \mathop = 1}^{k + 1} \left({x_i + y_i}\right)$

### Induction Step

This is the induction step:

\(\displaystyle \sum_{i \mathop = 1}^{k + 1} x_i + \sum_{i \mathop = 1}^{k + 1} y_i\) | \(=\) | \(\displaystyle \left({\sum_{i \mathop = 1}^k x_i + x_{k + 1} }\right) + \left({\sum_{i \mathop = 1}^k y_i + y_{k + 1} }\right)\) | Definition of Summation | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({\sum_{i \mathop = 1}^k x_i + \sum_{i \mathop = 1}^k y_i}\right) + \left({x_{k + 1} + y_{k + 1} }\right)\) | Commutative Law of Addition and Associative | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^k \left({x_i + y_i}\right) + \left({x_{k + 1} + y_{k + 1} }\right)\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^{k + 1} \left({x_i + y_i}\right)\) | Definition of Summation |

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

- $\displaystyle \forall n \in \N_{> 0}: \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \left({x_i + y_i}\right)$

$\blacksquare$