Answer
See explanation
Work Step by Step
Below is a standard induction proof, using the recursive definition of summation, the triangle inequality, and the definition of absolute value, to show that for all positive integers \(n\),
\[
\biggl|\;\sum_{i=1}^n a_i\biggr|
\;\le\;
\sum_{i=1}^n |a_i|,
\]
where \(a_1,\dots,a_n\) are real numbers.
---
## 1. Recursive Definition of Summation
Recall that for a positive integer \(n\),
\[
\sum_{i=1}^n a_i
\;=\;
\begin{cases}
a_1, & \text{if }n=1,\\
\Bigl(\sum_{i=1}^{n-1} a_i\Bigr) \;+\; a_n, & \text{if }n>1.
\end{cases}
\]
---
## 2. Proof by Mathematical Induction
### Base Case \((n=1)\)
When \(n=1\),
\[
\bigl|\sum_{i=1}^1 a_i\bigr| = |a_1|,
\quad
\sum_{i=1}^1 |a_i| = |a_1|.
\]
Hence
\[
\bigl|\sum_{i=1}^1 a_i\bigr| = |a_1| \le |a_1|,
\]
so the statement holds for \(n=1\).
### Inductive Step
Assume the statement holds for some positive integer \(n\). In other words, assume
\[
\biggl|\;\sum_{i=1}^n a_i\biggr|
\;\le\;
\sum_{i=1}^n |a_i|.
\]
We want to prove it then holds for \(n+1\). By the recursive definition of summation,
\[
\sum_{i=1}^{n+1} a_i
\;=\;
\Bigl(\sum_{i=1}^n a_i\Bigr) \;+\; a_{n+1}.
\]
Using the triangle inequality \(\,\bigl|x+y\bigr|\le|x|+|y|\), we get
\[
\biggl|\;\sum_{i=1}^{n+1} a_i\biggr|
\;=\;
\bigl|\bigl(\sum_{i=1}^n a_i\bigr) + a_{n+1}\bigr|
\;\le\;
\biggl|\sum_{i=1}^n a_i\biggr|\;+\;|a_{n+1}|.
\]
By the inductive hypothesis,
\[
\biggl|\sum_{i=1}^n a_i\biggr|
\;\le\;
\sum_{i=1}^n |a_i|.
\]
Hence,
\[
\biggl|\;\sum_{i=1}^{n+1} a_i\biggr|
\;\le\;
\sum_{i=1}^n |a_i|\;+\;|a_{n+1}|
\;=\;
\sum_{i=1}^{n+1} |a_i|.
\]
Thus the statement holds for \(n+1\). By mathematical induction, it is true for all positive integers \(n\).
---
## 3. Conclusion
We have shown, for all \(n \ge 1\),
\[
\boxed{
\biggl|\;\sum_{i=1}^n a_i\biggr|
\;\le\;
\sum_{i=1}^n |a_i|.
}
\]
This completes the proof via the recursive definition of summation, the triangle inequality, and induction.
