Product with Sum of Scalar

Theorem
Let $\left({G, +_G}\right)$ be an abelian group whose identity is $e$.

Let $\left({R, +_R, \times_R}\right)$ be a ring whose zero is $0_R$.

Let $\left({G, +_G, \circ}\right)_R$ be an $R$-module.

Let $x \in G, \lambda \in R$.

Let $\left \langle {\lambda_m} \right \rangle$ be a sequence of elements of $R$ i.e. scalars.

Then:
 * $\displaystyle \left({\sum_{k \mathop = 1}^m \lambda_k}\right) \circ x = \sum_{k \mathop = 1}^m \left({\lambda_k \circ x}\right)$

Proof
This follows by induction from Module: $(2)$, as follows.

For all $m \in \N^*$, let $P \left({m}\right)$ be the proposition:
 * $\displaystyle \left({\sum_{k \mathop = 1}^m \lambda_k}\right) \circ x = \sum_{k \mathop = 1}^m \left({\lambda_k \circ x}\right)$

Basis for the Induction
$P(1)$ is true, as this just says:
 * $\lambda_1 \circ x = \lambda_1 \circ x$

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({n}\right)$ is true, where $n \ge 1$, then it logically follows that $P \left({n+1}\right)$ is true.

So this is our induction hypothesis:


 * $\displaystyle \left({\sum_{k \mathop = 1}^n \lambda_k}\right) \circ x = \sum_{k \mathop = 1}^n \left({\lambda_k \circ x}\right)$

Then we need to show:


 * $\displaystyle \left({\sum_{k \mathop = 1}^{n+1} \lambda_k}\right) \circ x = \sum_{k \mathop = 1}^{n+1} \left({\lambda_k \circ x}\right)$

Induction Step
This is our induction step:

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

Therefore:
 * $\displaystyle \forall m \in \N^*: \left({\sum_{k \mathop = 1}^m \lambda_k}\right) \circ x = \sum_{k \mathop = 1}^m \left({\lambda_k \circ x}\right)$