Problem 11 - College Algebra

Question

Write the first four terms of each sequence whose general term is given. $$a_{n}=\frac{(-1)^{n+1}}{2^{n}-1}$$

Step-by-Step Solution

Answer

The first four terms of the sequence are -1, 1/3, -1/7, 1/15.

Step 1: Understanding the sequence

The given general term of the sequence is \(a_{n}=\frac{(-1)^{n+1}}{2^{n}-1}\). This means that for each term of the sequence, we substitute the term number into 'n' in the general term to get the value of the term.

Step 2: Getting the first term

To get the first term (n=1), we substitute n=1 into the general term to have \(a_{1} = \frac{(-1)^{1+1}}{2^{1}-1} = \frac{-1}{1}=-1\)

Step 3: Getting the second term

Similarly, we can get the second term by substituting n=2 into the general term: \(a_{2} = \frac{(-1)^{2+1}}{2^{2}-1} = \frac{1}{3}\)

Step 4: Getting the third term

Substituting n=3 into the general term gives us: \(a_{3} = \frac{(-1)^{3+1}}{2^{3}-1} = \frac{-1}{7}\)

Step 5: Getting the fourth term

Substitute n=4 into the general term to get: \(a_{4} = \frac{(-1)^{4+1}}{2^{4}-1} = \frac{1}{15}\)