Question

Write an expression for the \(n\) th term of the sequence. (There is more than one correct answer.) \(1,-\frac{1}{1 \cdot 3}, \frac{1}{1 \cdot 3 \cdot 5},-\frac{1}{1 \cdot 3 \cdot 5 \cdot 7}, \ldots\)

Short answer

The nth term of the sequence can be expressed as \(a_n = (-1)^{n+1} \cdot \frac{1}{(2n-1)!!}\)

Step-by-step solution

Identify the sequence pattern

Look at the pattern given. Notice that the denominators are increasing products of odd numbers, and the signs of the terms alternate. All denominators have odd numbers and there is no common difference. This suggests that the expression involves factorials or a variation of factorial.

Express the pattern using mathematical notation

Given that if n = 1, we start with 1, and then start multiplying by successive odd numbers as n increases. We can write the denominator as the product of first n odd numbers. The alternating sign can be expressed as \(-1^{n+1}\). The term is then expressed as \(a_n= (-1)^{n+1} \cdot \frac{1}{(2n-1)!!}\) where \((2n-1)!!\) represents the double factorial of odd numbers, which means multiplying together the sequence of numbers with a common difference of 2. If n=1, there is no denominator which aligns with our sequence.

Verification

Verify the formula by substituting the first few values of n. If the terms match with the given sequence, then the formula for the nth term is correct.