# Summation is Linear/Scaling of Summations

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## 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:

- $\displaystyle \lambda \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda x_i$

## Proof

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

- $\displaystyle \lambda \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda x_i$

### Basis for the Induction

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

\(\displaystyle \lambda \sum_{i \mathop = 1}^1 x_i\) | \(=\) | \(\displaystyle \lambda x_1\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^1 \lambda x_i\) |

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:

- $\displaystyle \lambda \sum_{i \mathop = 1}^k x_i = \sum_{i \mathop = 1}^k \lambda x_i$

from which it is to be shown that:

- $\displaystyle \lambda \sum_{i \mathop = 1}^{k + 1} x_i = \sum_{i \mathop = 1}^{k + 1} \lambda x_i$

### Induction Step

This is the induction step:

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

\(\displaystyle \) | \(=\) | \(\displaystyle \lambda \sum_{i \mathop = 1}^k x_i + \lambda x_{k + 1}\) | Multiplication of Numbers Distributes over Addition | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^k \lambda x_i + \lambda x_{k + 1}\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^{k + 1} \lambda x_i\) | 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:

- $\displaystyle \forall n \in \N_{> 0}: \lambda \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda x_i$

$\blacksquare$