Summation is Linear/Scaling of Summations

From ProofWiki
Jump to navigation Jump to search


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.


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


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.


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