Problem 59 - Algebra 2

Question

What is a recursive formula for the sequence whose explicit formula is \(a_{n}=(n+1)^{2} ?\) F. \(a_{1}=1, a_{n}=\left(a_{n-1}+1\right)^{2}\) H. \(a_{1}=n, a_{n}=a_{n-1}+n\) G. \(a_{1}=4, a_{n}=\left(\sqrt{a_{n-1}}+1\right)^{2}\) J. \(a_{1}=n^{2}, a_{n}=\left(a_{n}-1\right)^{2}+1\)

Step-by-Step Solution

Answer

The recursive formula for the sequence whose explicit formula is \(a_{n}=(n+1)^{2}\) is G. \(a_{1}=4, a_{n}=\left(\sqrt{a_{n-1}}+1\right)^{2}\).

Step 1: Understand the explicit formula

The explicit formula given is \(a_{n}=(n+1)^{2}\). This means each nth term in the sequence is squared of one more than n. So, \(a_{1}=(1+1)^{2}=4\), \(a_{2}=(2+1)^{2}=9\), \(a_{3}=(3+1)^{2}=16\) and so on.

Step 2: Review the recursive options given

Now we need to see which of the recursive rules matches the pattern of the sequence we've obtained. We try all the given options: \n Option F: \(a_{1}=1, a_{n}=\left(a_{n-1}+1\right)^{2}\). Here, square of one more than the previous term is used. But the starting value doesn't match with the one from explicit formula (4), so this can't be the recursion rule.\n Option H: \(a_{1}=n, a_{n}=a_{n-1}+n\). This rule adds the term number to the previous term. This doesn't match the sequence generated by explicit formula, so this also can't be the rule.\n Option G: \(a_{1}=4, a_{n}=\left(\sqrt{a_{n-1}}+1\right)^{2}\). This takes square of one more than the square-root of the previous term. Also, the starting value is the one we have from explicit formula, which is 4. So this might be the correct recursive formula.\n Option J: \(a_{1}=n^{2}, a_{n}=\left(a_{n}-1\right)^{2}+1\). Here too the starting value doesn't match and the rule is not matching the sequence obtained by explicit formula. So this can't be the rule.

Step 3: Check the possible recursion formula

The possible recursion formula is \(a_{1}=4, a_{n}=\left(\sqrt{a_{n-1}}+1\right)^{2}\). Let's check if this yields the same sequence as obtained by explicit formula: \(a_{1}=4\), \(a_{2}=\left(\sqrt{4}+1\right)^{2}=9\), \(a_{3}=\left(\sqrt{9}+1\right)^{2}=16\). So, this matches the sequence pattern of the explicit formula.

Key Concepts

Explicit Formula

An explicit formula in mathematics allows you to find the nth term of a sequence directly. It gives you a way to compute the value without needing the preceding terms. For example, in the sequence represented by the explicit formula \(a_n = (n+1)^2\), this means that the nth term is the square of one more than \(n\).
This is a powerful tool when you want to quickly calculate or verify a term in the sequence without going through each previous term. In this case, you start by plugging in the value of \(n\) into the formula. For \(n = 1\), \(a_1 = (1+1)^2 = 4\); for \(n = 2\), \(a_2 = (2+1)^2 = 9\), and so on.

• Saves time by directly providing the term
• Provides a clear pattern or rule for the sequence

Understanding the explicit formula helps ensure clarity and efficient calculations, especially where sequences become complex or lengthy.

Sequences in Mathematics

Sequences are ordered lists of numbers that follow a particular pattern or rule. In mathematics, these patterns can be identified using formulas. Two common types of formulas used to describe sequences are explicit and recursive formulas.

• An explicit formula allows for direct computation of any term in the sequence as discussed earlier.
• A recursive formula provides a way to find a term based on the previous term(s).

Sometimes, identifying these rules can be like solving a puzzle. You need to notice how each term is derived from its predecessor or from a given rule.
In the case of our example, sequence representations clarify differences. The explicit formula \(a_n = (n+1)^2\) gives you a straightforward approach to each term. However, you might use a recursive formula, like \(a_1=4, a_{n}=(\sqrt{a_{n-1}}+1)^2\), for iteration by building from previous terms.

Square Root

The square root is a fundamental concept in mathematics. It involves finding a number which, when multiplied by itself, gives the original number. For example, the square root of 9 is 3, since \(3 \times 3 = 9\).
In our sequence problem, the recursive formula \(a_{n}=\left(\sqrt{a_{n-1}}+1\right)^{2}\) incorporates the square root to define progression through terms. By taking the square root of the previous term and adding 1, before squaring again, ensures that the recursive formula generates the same sequence as the explicit formula in our example.

• This demonstrates a relationship between square roots and sequence generation.
• Square roots are often used to simplify expressions or revert a square operation.

Such operations are common, especially in sequences where either growth or reduction patterns rely on roots and squares to maintain relationships between terms.