CASE 2: \( r<-1 \) 1.3.1 If \( r=-1,001 \), determine the value of \( r^{a} \) if (a) \( n=9999 \) (b) \( n=10000 \) 1.3.2 Make a conjecture with regard to \( r^{n} \) and \( S_{n} \) as \( n \rightarrow \infty \)
Upstudy AI Solution
Answer
Solution

Sign in to Unlock Answers for Free!
A Learning Platform Trusted by Millions of Real Students and Teachers.

Answered by UpStudy AI and reviewed by a Professional Tutor
Mind Expander
When \( r = -1.001 \), calculating \( r^n \) for large \( n \) shows some interesting behavior. (a) For \( n = 9999 \), \( r^{9999} \) will be a very large negative number because its absolute value is greater than 1, meaning it will grow in magnitude as \( n \) increases. (b) For \( n = 10000 \), \( r^{10000} \) becomes a very large positive number since we’ve raised a negative base to an even power, flipping the sign. As \( n \) approaches infinity, our conjecture for \( r^n \) suggests that the magnitude of \( r^n \) will grow unbounded, alternating between negative and positive based on whether \( n \) is odd or even. Therefore, \( S_n \) (the sum of a geometric series) may not converge but rather oscillate or diverge, reflecting the growing values of \( r^n \).