Summation is Linear/Scaling 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:

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


Proof

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

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


Basis for the Induction

$\map P 1$ is the case:

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

This is the basis for the induction.


Induction Hypothesis

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


So this is the induction hypothesis:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:

\(\ds \lambda \sum_{i \mathop = 1}^{k + 1} x_i\) \(=\) \(\ds \lambda \paren {\sum_{i \mathop = 1}^k x_i + x_{k + 1} }\) Definition of Summation
\(\ds \) \(=\) \(\ds \lambda \sum_{i \mathop = 1}^k x_i + \lambda x_{k + 1}\) Multiplication of Numbers Distributes over Addition
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^k \lambda x_i + \lambda x_{k + 1}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^{k + 1} \lambda x_i\) Definition of Summation

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


Therefore:

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

$\blacksquare$