Proof That ∑cai = c ∑ai Where c Is A Constant (Distributivity)

Consider the summation:

\[ \sum_{i=1}^n ca_i = ca_1 + ca_2 + \cdots + ca_{n-1} + ca_n \]

We can use the distributive property of multiplication:

\[\begin{align} \sum_{i=1}^n ca_i &= c(a_1 + a_2 + \cdots + a_{n-1} + a_n) \\ &= c \left( \sum_{i=1}^n a_i \right) \end{align}\]

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