Problem 10 - Calculus for Biology and Medicine

Question

Determine the values of the sequence \(\left\\{a_{n}\right\\}\) for \(n=0,1,2, \ldots, 5\) $$ a_{n}=\frac{(-1)^{n}}{(n+1)^{2}} $$

Step-by-Step Solution

Answer

The sequence values are: 1, -0.25, 0.111, -0.0625, 0.04, -0.0278.

Step 1: Calculate Value for n=0

Substitute \(n=0\) into the formula \(a_n = \frac{(-1)^n}{(n+1)^2}\). This becomes: \[ a_0 = \frac{(-1)^0}{(0+1)^2} = \frac{1}{1} = 1. \]

Step 2: Calculate Value for n=1

Substitute \(n=1\) into the formula. This becomes: \[ a_1 = \frac{(-1)^1}{(1+1)^2} = \frac{-1}{4} = -0.25. \]

Step 3: Calculate Value for n=2

Substitute \(n=2\) into the formula. This calculation is: \[ a_2 = \frac{(-1)^2}{(2+1)^2} = \frac{1}{9} \approx 0.111. \]

Step 4: Calculate Value for n=3

Substitute \(n=3\) into the formula. This becomes: \[ a_3 = \frac{(-1)^3}{(3+1)^2} = \frac{-1}{16} = -0.0625. \]

Step 5: Calculate Value for n=4

Substitute \(n=4\) into the formula. This is: \[ a_4 = \frac{(-1)^4}{(4+1)^2} = \frac{1}{25} = 0.04. \]

Step 6: Calculate Value for n=5

Substitute \(n=5\) into the formula. This becomes: \[ a_5 = \frac{(-1)^5}{(5+1)^2} = \frac{-1}{36} \approx -0.0278. \]

Key Concepts

Understanding Alternating Series

In mathematics, a sequence that changes sign across its terms is known as an alternating series. This concept is vital when dealing with sequences like the given \(a_n = \frac{(-1)^n}{(n+1)^2}\). The term 'alternating' comes from the sequence flipping between positive and negative values as we move from one term to the next.
An alternating series can often be identified by the presence of \( (-1)^n \) or \( (-1)^{n+1} \) in the sequence formula. Let's see how it works in our sequence:

• For even values of n (0, 2, 4,...), \( (-1)^n \) becomes 1, resulting in a positive term.
• For odd values of n (1, 3, 5,...), \( (-1)^n \) turns into -1, creating a negative term.

This alternating fashion is crucial because it influences the convergence and behavior of the series over its domain. As you progress through the values of n, you observe that the terms alternate in sign, helping understand the series' pattern.

Demystifying Index Calculation

Index calculation involves evaluating the sequence at specific positions by substituting different index values into its formula. In the problem provided, you're given \( n \) to be the index, which ranges from 0 to 5. Let's see how this unfolds in our sequence case:
The key to solving this lies in simply substituting each index value directly into the given sequence formula. This means each value of \( n \) requires its own calculation by following the structure of \( a_n = \frac{(-1)^n}{(n+1)^2} \).

• Start with \( n = 0 \), substitute it into the formula, and solve it to find \( a_0 \).
• Repeat the process for \( n = 1, 2, 3, 4, \) and \( 5 \) to determine each respective term \( a_1, a_2, \) etc.

The index value is essential as it also decides the sign changes, owing to the \( (-1)^n \) factor. Practicing index calculation gives a clear picture of each term's contribution in the bigger sequence.

Exploring the Substitution Method

The substitution method is a powerful technique used to compute terms of sequences or functions efficiently. In our focused sequence exercise, substitution plays a critical role in simplifying calculations.
Substitution involves replacing the variable \( n \) in a sequence formula with specific values to compute each term:

• Take the given sequence formula: \( a_n = \frac{(-1)^n}{(n+1)^2} \).
• Substitute \( n = 0 \) into the formula to find \( a_0 \).
• Continue this substitution process for \( n = 1, 2, \) and so on up to 5, making sure to calculate \( a_n \) individually for each value.

This method simplifies what could be a complex arithmetic task, turning it into a straightforward step-by-step process. The substitution approach is especially useful when working with sequences that involve expressions changing due to varying index values. It efficiently translates mathematical expressions into tangible results, building a deeper understanding of sequences.