Problem 9 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

All values \( \left| a_n \right| \) from \( n=0 \) to 5 are 0.

Step 1: Calculate for n=0

For the sequence, when \( n = 0 \), calculate \( a_0 \). To do this, evaluate:\[a_0 = (-1)^0 + (-1)^{0+1} = 1 + (-1) = 0\]Thus, \( \left| a_0 \right| = |0| = 0 \).

Step 2: Calculate for n=1

Next, calculate \( a_1 \) by plugging in \( n = 1 \):\[a_1 = (-1)^1 + (-1)^{1+1} = -1 + 1 = 0\]Therefore, \( \left| a_1 \right| = |0| = 0 \).

Step 3: Calculate for n=2

For \( n = 2 \), evaluate \( a_2 \):\[a_2 = (-1)^2 + (-1)^{2+1} = 1 + (-1) = 0\]Thus, \( \left| a_2 \right| = |0| = 0 \).

Step 4: Calculate for n=3

Calculate when \( n = 3 \):\[a_3 = (-1)^3 + (-1)^{3+1} = -1 + 1 = 0\]So, \( \left| a_3 \right| = |0| = 0 \).

Step 5: Calculate for n=4

For \( n = 4 \), determine \( a_4 \):\[a_4 = (-1)^4 + (-1)^{4+1} = 1 + (-1) = 0\]Hence, \( \left| a_4 \right| = |0| = 0 \).

Step 6: Calculate for n=5

Finally, calculate with \( n = 5 \):\[a_5 = (-1)^5 + (-1)^{5+1} = -1 + 1 = 0\]Thus, \( \left| a_5 \right| = |0| = 0 \).

Key Concepts

Absolute Value

The absolute value is a fundamental concept in mathematics. It represents the non-negative value of a number without regard to its sign. So, for any real number, its absolute value is always zero or positive because it measures the distance of that number from zero on a number line. If you think about it like this: whether a number is positive or negative, what matters is how far it is from zero.

To denote the absolute value, we use vertical bars: for a number \( x \), the absolute value is written as \( |x| \). This means:

• \( |x| = x \) if \( x \geq 0 \)
• \( |x| = -x \) if \( x < 0 \)

In the context of the sequence \( |a_n| \) from our problem, all computed values for \( a_n \) resulted in zero. Consequently, the sequence \( |a_n| \), which is simply the absolute values of the sequence terms, also carries zero for all tested \( n \).

This reflects an important property of absolute values, where even if the term itself was negative, the absolute value would give us a non-negative result.

Alternating Sequence

An alternating sequence is a type of sequence in which the signs of the elements switch back and forth between positive and negative. This distinct pattern can create an interesting rhythm to the numerical sequence. In mathematical terms, an alternating sequence can be represented by alternating powers of \(-1\).

In the given problem, the sequence was defined using the formula \( a_n = (-1)^n + (-1)^{n+1} \). Here:

• The term \((-1)^n\) depends on whether \(n\) is odd or even.
• When \(n\) is even, \((-1)^n = 1\) and \((-1)^{n+1} = -1\), leading the sum to zero.
• When \(n\) is odd, \((-1)^n = -1\) and \((-1)^{n+1} = 1\), again resulting in a sum of zero.

Hence, even though each component of the sequence potentially changes signs, the overall result of each alternate pair of terms cancels each other out, leading to a sequence whose every \(a_n\) becomes zero for all values of \(n\). This showcases an interesting property of how alternating sequences can sometimes harmonize to a constant outcome.

Integer Sequence

An integer sequence is a sequence where each term is an integer. Sequences can be finite or infinite and often follow specific rules or formulas to determine successive terms. In our problem, the sequence \(a_n\) is crafted using exponents of \(-1\), which are integer-valued operations.

When the value of \((-1)^n + (-1)^{n+1}\) was evaluated for each \(n\) from 0 to 5, it resulted in terms like 0, which are integers. An integer has no fractional or decimal parts, making them clean, whole numbers. This quality of integer sequences means they are often easier to handle and calculate.

By understanding that the operation involved produces integers, students can grasp that all terms \(a_n\) will reliably be zero, reinforcing the concrete nature of integer sequences. The simplicity of the resulting terms—combined with the intriguing behavior of \(-1\)—brings a satisfying closure to observing integer sequences, especially those involving alternating terms.