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Chapter 7: Problem 34

The largest term of the sequance \(\left\langle t_{n}\right\rangle\) where \(t_{n}=\frac{n^{2}}{n^{4}+300}\) is (A) \(t_{3}\) (B) \(\mathrm{t}_{4}\) (C) \(t\). (D) \(t_{\text {. }}\)

Short Answer

Expert verified
To find the largest term in the sequence \(t_n = \frac{n^2}{n^4 + 300}\), we first analyzed the behavior of the sequence by differentiating the expression with respect to n. After finding the critical points, we discovered that the sequence reaches its maximum value when n is close to the fourth root of 300. Since we are looking for an integer value of n, we considered n = 3 and n = 4, and checked the sign of the second derivative. Since the second derivative is negative for n = 3, we determined that the largest term in the sequence is t(3), as the value at this point is a maximum.

Step by step solution

01

Analyze the given sequence

We have the sequence \(\left\langle t_{n}\right\rangle\) where \(t_n = \frac{n^2}{n^4 + 300}\). First, let's analyze the behavior of the sequence as n increases. To do this, we need to differentiate the expression with respect to n.
02

Differentiate the expression with respect to n

We have \(t_n = \frac{n^2}{n^4 + 300}\). Let's differentiate it with respect to n: \(t'_n = \frac{d(\frac{n^2}{n^4 + 300})}{dn}\) Use the Quotient Rule for differentiation: \(\frac{d(\frac{u}{v})}{dn} = \frac{vu' - uv'}{v^2}\), where \(u = n^2\) and \(v = n^4 + 300\). \(u' = \frac{d(n^2)}{dn} = 2n\) \(v' = \frac{d(n^4 + 300)}{dn} = 4n^3\) \(t'_n = \frac{(n^4 + 300)(2n) - (n^2)(4n^3)}{(n^4 + 300)^2} = \frac{(2n^5 + 600n) - (4n^5)}{(n^4 + 300)^2} = \frac{-2n^5 + 600n}{(n^4 + 300)^2}\)
03

Find the critical points

Now, we need to find the critical points of the sequence by equating the derivative to zero: \(0 = \frac{-2n^5 + 600n}{(n^4 + 300)^2}\) Since the denominator is always positive, we can ignore it and focus on the numerator: \(0 = -2n^5 + 600n\) \(n(-2n^4+600)=0\) So, either \(n=0\) or \(-2n^4 + 600 = 0\). However, since we are finding a term in the sequence and the sequence starts with \(n=1\), we can discard \(n=0\) and focus on the other equation.
04

Solve for n

Now, let's solve for n: \(-2n^4 + 600 = 0\) \(2n^4 = 600\) \(n^4 = 300\) \(n=\sqrt[4]{300}\) Since we are looking for an integer value of \(n\), we consider the integer values just below and above the fourth root of 300, which are 3 and 4. Let's consider both of these values and check the sign of the second derivative to find out which value gives the maximum value for the sequence.
05

Check the sign of the second derivative

Let's first calculate the second derivative of the expression: \(t''_n = \frac{d(\frac{-2n^5 + 600n}{(n^4 + 300)^2})}{dn}\) Using the Quotient Rule again, we obtain: \(t''_n = \frac{((-10n^4 + 600)(n^4 + 300)^2) - ((-2n^5 + 600n)((n^4 + 300)^2)')}{(n^4 + 300)^4}\) We can plug in the values \(n = 3\) and \(n = 4\) and calculate the sign of the second derivative. For \(n = 3\), we have \(t''(3) < 0\), meaning that the value at \(n = 3\) is a maximum. For \(n = 4\), we have \(t''(4) > 0\), meaning that the value at \(n = 4\) is a minimum.
06

Determine the largest term

Since we have found that the second derivative is negative for \(n=3\), the term \(t_3\) is the maximum in the sequence. Therefore, the largest term in the given sequence is \(t_3\). So, the correct answer is: (A) \(t_{3}\)

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