Answer
See explanation
Work Step by Step
Below is a standard proof by induction (using the recursive definition of summation) of the generalized distributive law:
**Claim.** For any positive integer \(n\) and any real numbers \(a_1,\dots,a_n\) and \(c\),
\[
\sum_{i=1}^n \bigl(c\,a_i\bigr) \;=\; c \,\sum_{i=1}^n a_i.
\]
---
## 1. Recursive Definition of Summation
We define, for a positive integer \(n\):
\[
\sum_{i=1}^n a_i
\;=\;
\begin{cases}
a_1, & \text{if }n=1,\\[6pt]
\Bigl(\sum_{i=1}^{n-1} a_i\Bigr) + a_n, & \text{if }n>1.
\end{cases}
\]
---
## 2. Proof by Induction
### **Base Case** \((n=1)\)
For \(n=1\), we have
\[
\sum_{i=1}^1 \bigl(c\,a_i\bigr)
= c\,a_1.
\]
Also,
\[
c\,\sum_{i=1}^1 a_i
= c\,a_1.
\]
Thus both sides agree (both are \(c\,a_1\)), so the statement holds for \(n=1\).
### **Inductive Step**
Assume the statement holds for some positive integer \(n\). In other words, assume
\[
\sum_{i=1}^n \bigl(c\,a_i\bigr)
\;=\;
c\,\sum_{i=1}^n a_i.
\]
We must prove it then holds for \(n+1\).
Using the recursive definition of summation for \(n+1\):
\[
\sum_{i=1}^{n+1} \bigl(c\,a_i\bigr)
\;=\;
\Bigl(\sum_{i=1}^{n} \bigl(c\,a_i\bigr)\Bigr) + \bigl(c\,a_{n+1}\bigr).
\]
By the inductive hypothesis,
\[
\sum_{i=1}^{n} \bigl(c\,a_i\bigr)
= c\,\sum_{i=1}^n a_i.
\]
Hence,
\[
\sum_{i=1}^{n+1} \bigl(c\,a_i\bigr)
= c\,\sum_{i=1}^n a_i \;+\; c\,a_{n+1}
= c\,\Bigl(\sum_{i=1}^n a_i + a_{n+1}\Bigr)
= c\,\sum_{i=1}^{n+1} a_i,
\]
where in the last step we again used the recursive definition of summation for \(\sum_{i=1}^{n+1} a_i\).
Thus the statement is true for \(n+1\). By the principle of mathematical induction, the equality
\[
\sum_{i=1}^n (c\,a_i) = c\,\sum_{i=1}^n a_i
\]
holds for all positive integers \(n\).
---
## 3. Conclusion
This completes the proof that, for all \(n \ge 1\),
\[
\sum_{i=1}^n \bigl(c\,a_i\bigr) \;=\; c \,\sum_{i=1}^n a_i,
\]
using the recursive definition of summation and mathematical induction.
