# Summation is Linear/Sum of Summations

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## Theorem

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

Let $\lambda$ be a number.

Then:

- $\ds \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \paren {x_i + y_i}$

## Proof

The proof proceeds by mathematical induction.

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

- $\ds \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \paren {x_i + y_i}$

### Basis for the Induction

$\map P 1$ is the case:

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

\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^1 \paren {x_i + y_i}\) |

This is the basis for the induction.

### Induction Hypothesis

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

So this is the induction hypothesis:

- $\ds \sum_{i \mathop = 1}^k x_i + \sum_{i \mathop = 1}^k y_i = \sum_{i \mathop = 1}^k \paren {x_i + y_i}$

from which it is to be shown that:

- $\ds \sum_{i \mathop = 1}^{k + 1} x_i + \sum_{i \mathop = 1}^{k + 1} y_i = \sum_{i \mathop = 1}^{k + 1} \paren {x_i + y_i}$

### Induction Step

This is the induction step:

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

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

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

\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^{k + 1} \paren {x_i + y_i}\) | Definition of Summation |

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

Therefore:

- $\ds \forall n \in \N_{> 0}: \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \paren {x_i + y_i}$

$\blacksquare$