Problem 20 - Calculus for Biology and Medicine

Question

Find the next four values of the sequence \(\left\\{a_{n}\right\\}\) on the basis of the values of \(a_{0}, a_{1}, a_{2}, a_{3}, a_{4}\). $$ -1, \frac{1}{4},-\frac{1}{9}, \frac{1}{16},-\frac{1}{25} $$

Step-by-Step Solution

Answer

The next four values are: \(-\frac{1}{36}, \frac{1}{49}, -\frac{1}{64}, \frac{1}{81}\).

Step 1: Identify the Pattern

Observe the sequence provided: -1, \( \frac{1}{4} \), \( -\frac{1}{9} \), \( \frac{1}{16} \), \( -\frac{1}{25} \). The sequence alternates between negative and positive values. The denominators are perfect squares: \(1^2, 2^2, 3^2, 4^2, 5^2\). The numerator alternately uses -1 and 1.

Step 2: Express the General Formula

Considering the pattern observed, express the sequence formula: \( a_n = \frac{(-1)^{n}}{(n+1)^2} \). This formula captures both the alternating sign due to \((-1)^n\) and the perfect square denominators with \((n+1)^2\).

Step 3: Calculate Next Sequence Value a5

Substitute \(n=5\) in the formula: \(a_5 = \frac{(-1)^5}{(5+1)^2} = \frac{-1}{6^2} = -\frac{1}{36}\).

Step 4: Calculate Next Sequence Value a6

Substitute \(n=6\) in the formula: \(a_6 = \frac{(-1)^6}{(6+1)^2} = \frac{1}{7^2} = \frac{1}{49}\).

Step 5: Calculate Next Sequence Value a7

Substitute \(n=7\) in the formula: \(a_7 = \frac{(-1)^7}{(7+1)^2} = \frac{-1}{8^2} = -\frac{1}{64}\).

Step 6: Calculate Next Sequence Value a8

Substitute \(n=8\) in the formula: \(a_8 = \frac{(-1)^8}{(8+1)^2} = \frac{1}{9^2} = \frac{1}{81}\).

Key Concepts

Alternating Series

An alternating series is a fascinating type of mathematical sequence where the signs of the terms alternate between positive and negative. This can be seen in the sequence provided, as it switches signs with each term:

• The terms alternate due to the factor \((-1)^n\) in the general formula. This mathematical expression is responsible for changing the sign of a term based on whether \(n\) is odd or even.
• For instance, when \(n\) is odd, \((-1)^n\) will result in a negative term, while an even \(n\) will produce a positive term. This pattern creates a regular oscillation of signs in the sequence.

Recognizing an alternating pattern can help predict the behavior and characteristics of a sequence. This knowledge is not only useful in identifying sequence trends but also in understanding convergence, especially when dealing with infinite series.

Perfect Squares

Perfect squares are an important concept in sequences and are simply squares of whole numbers. They appear prominently in the sequence discussed, specifically in the denominator of the terms:

• The denominators of the given sequence are 1, 4, 9, 16, 25, etc., which are squares of the numbers 1, 2, 3, 4, and 5.
• This is represented in our sequence's general formula by the term \((n+1)^2\), where the index \(n\) starts from 0. The "+1" adjustment shifts the sequence of numbers to start from the perfect square of 1 instead of 0.

Identifying perfect squares within sequences is essential for recognizing patterns and has applications in solving various mathematical problems. Understanding how to express numbers as perfect squares allows for easing calculations and simplifying complex expressions.

General Formula in Sequences

The general formula is a concise expression to define a sequence and determine its terms. Here, the formula \(a_n = \frac{(-1)^n}{(n+1)^2}\) incorporates two essential concepts:

• The term \((-1)^n\) handles the alternating signs of the sequence, efficiently flipping from positive to negative, ensuring the alternation pattern is maintained throughout.
• The expression \((n+1)^2\) establishes the perfect square pattern in the sequence's denominator, ensuring each denominator is the square of a whole number.

Utilizing a general formula not only simplifies the process of calculating specific terms but also unveils the inherent structure within a sequence. Formulas like these are critical in broader mathematical analysis, as they allow for quick identification of series and understanding their properties without listing every term individually.