Summation is Linear

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 \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n

\left({x_i + y_i}\right)$
 * $\displaystyle \lambda \cdot \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda \cdot x_i$

Proof of the first statement
Let $P_1$(n) be the predicate indicating that $\displaystyle \sum_{i \mathop = 1}^n x_i + \sum_{i \mathop = 1}^n y_i = \sum_{i \mathop = 1}^n \left({x_i + y_i}\right)$ is true.

Base Case
$P_1$(1):


 * $\because L.H.S. \mathop = R.H.S.$
 * $\therefore P_1$(1) is true.

Induction Step
Assume that $P_1$(k) is true.

$P_1$(k+1):


 * $\because L.H.S. \mathop = R.H.S.$
 * $\therefore P_1$(k+1) is true.

Therefore, we can conclude that the first statement of this theory is true.

Proof of the second statement
Let $P_2$(n) be the predicate indicating that $\displaystyle \lambda \cdot \sum_{i \mathop = 1}^n x_i = \sum_{i \mathop = 1}^n \lambda \cdot x_i$ is true.

Base Case
$P_2$(1):


 * $\because L.H.S. \mathop = R.H.S$
 * $\therefore P_2$(1) is true.

Induction Step
Assume that $P_2$(k) is true.

$P_2$(k+1):


 * $\because L.H.S. \mathop = R.H.S.$
 * $\therefore P_2$(k+1) is true.

Therefore, we can conclude that the second statement of this theory is true.