Problem 80 - Algebra and Trigonometry

Question

Different Sequences That Start the Same (a) Show that the first four terms of the sequence \(a_{n}=n^{2}\) are $$1,4,9,16, \ldots$$ (b) Show that the first four terms of the sequence \(a_{n}=n^{2}+(n-1)(n-2)(n-3)(n-4)\) are also $$1,4,9,16, \ldots$$ (c) Find a sequence whose first six terms are the same as those of \(a_{n}=n^{2}\) but whose succeeding terms differ from this sequence. (d) Find two different sequences that begin $$2,4,8,16, \ldots$$

Step-by-Step Solution

Answer

Sequences share first terms, e.g., quadratic sequence's structure gives identical initial terms. Further conditions produce same start with divergence, as demonstrated.

Step 1: Finding the First Terms of Sequence (a)

Calculate the first four terms of the sequence \(a_{n}=n^{2}\). Evaluate for \(n=1\), \(n=2\), \(n=3\), and \(n=4\). For \(n=1\), \(a_{1}=1^{2}=1\). For \(n=2\), \(a_{2}=2^{2}=4\). For \(n=3\), \(a_{3}=3^{2}=9\). Finally, for \(n=4\), \(a_{4}=4^{2}=16\). The sequence starts with \(1, 4, 9, 16, \ldots\).

Step 2: Finding the First Terms of Sequence (b)

Evaluate the first four terms of the sequence \(a_{n}=n^{2}+(n-1)(n-2)(n-3)(n-4)\). For \(n=1\), \(a_{1}=1^{2}+(1-1)(1-2)(1-3)(1-4)=1\). For \(n=2\), \(a_{2}=2^{2}+(2-1)(2-2)(2-3)(2-4)=4\). For \(n=3\), \(a_{3}=3^{2}+(3-1)(3-2)(3-3)(3-4)=9\). For \(n=4\), \(a_{4}=4^{2}+(4-1)(4-2)(4-3)(4-4)=16\). These terms match those of the sequence \(n^2\).

Step 3: Constructing Sequence (c)

To find a sequence whose first six terms match \(a_{n}=n^{2}\), consider \(a_{n}=n^{2}+(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)\). Evaluate the first six terms: for each \(n = 1, 2, 3, 4, 5, 6\), the second term becomes zero. For \(n=7\), calculate: \(a_{7}=7^{2}+(7-1)(7-2)(7-3)(7-4)(7-5)(7-6)=49+720=769\). Thus, the sequence shifts at the 7th term.

Step 4: Constructing Sequence (d) - First Sequence

To create a sequence starting with \(2, 4, 8, 16, \ldots\), we can use \(a_{n}=2^{n}\). Verify: For \(n=1\), \(a_{1}=2^{1}=2\); for \(n=2\), \(a_{2}=2^{2}=4\); for \(n=3\), \(a_{3}=2^{3}=8\); and for \(n=4\), \(a_{4}=2^{4}=16\).

Step 5: Constructing Sequence (d) - Second Sequence

Consider the sequence \(b_{n} = n + (n-1)\times(n-2)\times(n-3)\times(n-4)\). Calculate for validation: For \(n=1\), \(b_{1} = 1 + 0 = 1\) (choose 2 so that it matches starting terms); adjust to start with 2 for desired results with adjustment: \(b_{n}=2^{n}\).

Key Concepts

Quadratic Sequences

Quadratic sequences are mathematical sequences where the difference between consecutive terms forms a linear sequence. A quadratic sequence is typically represented by a formula of the form \( a_n = an^2 + bn + c \). Here, the term \( n^2 \) characterizes its quadratic nature.

To determine if a sequence is quadratic, examine the differences between consecutive terms:

• Calculate the first differences by subtracting each term from the next one.
• Find the second differences by subtracting consecutive first differences.

In quadratic sequences, the second differences are constant. This pattern helps predict values and understand the nature of the sequence.

In the example provided, the sequence \( a_n = n^2 \) yields initial terms of 1, 4, 9, 16, where the second differences are always 2, confirming its quadratic nature.

Polynomial Sequences

Polynomial sequences extend beyond simple quadratic form and include sequences defined by polynomials of any degree. Polynomial sequences can be expressed as \( a_n = a_k n^k + a_{k-1} n^{k-1} + \ldots + a_1 n + a_0 \). Here, \( k \) represents the highest degree of \( n \).

With polynomial sequences, the nature and pattern of the differences vary based on the degree of the polynomial. As the degree increases, so does the complexity.

In our example, the sequence \( a_n = n^2 + (n-1)(n-2)(n-3)(n-4) \) technically contains a polynomial because of the cubic term, but its first few terms mimic a quadratic sequence. It complicates the sequence but aligns initially with \( n^2 \) due to the factor making the added term zero for \( n = 1 \) to \( 4 \). This demonstrates how creative modifications can sustain appearances across sequence categories.

Exponential Sequences

Exponential sequences differ significantly from quadratic and polynomial sequences by growing rapidly. These sequences typically follow a pattern defined by \( a_n = a \cdot b^n \), where \( b \) is a constant base determining growth rate, and \( a \) is the initial term coefficient.

The primary trait of exponential sequences is their rate of increase – the terms grow by a multiplicative factor rather than an additive or polynomial pattern.

• Each term is \( b \) times the previous term.
• Growth accelerates with each step, often depicted in phenomena like population growth or compound interest.

In the exercise, the sequence \( a_n = 2^n \) starts with 2, 4, 8, 16, explicitly demonstrating this rapid growth. The base, 2, doubles each term, a hallmark of exponential behavior compared to the more gradual increases of quadratic or polynomial sequences.