Jump to content

Science:Math Exam Resources/Courses/MATH220/April 2005/Question 05/Solution 1

From UBC Wiki

In the base case when n=1, we have |a1|≤|a1|, so the statement is true.

For the inductive step, we assume that |∑i=1nai|≤∑i=1n|ai| holds for some (fixed) n∈ℕ. Then we show that it also holds with n replaced by n+1:

|∑i=1n+1ai|=|∑i=1nai+an+1|≤|∑i=1nai|+|an+1|≤∑i=1n|ai|+|an+1|=∑i=1n+1|ai|.

The first inequality is true by the triangle inequality and the second by the induction hypothesis. This finishes the inductive step and so the result is true for all n∈ℕ by induction.