Question

Let {𝑎_𝑛 }_𝑛≥1 be a sequence such that 𝑎₁=1 and 4𝑎_𝑛+1=\(\sqrt{45 + 16𝑎_𝑛} ,𝑛\)=1, 2, 3, … . Then, which one of the following statements is TRUE?

• A. {𝑎_𝑛 }_𝑛≥1 is monotonically increasing and converges to \(\frac{17}{8}\)
• B. {𝑎_𝑛 }_𝑛≥1 is monotonically increasing and converges to \(\frac{9}{4}\)
• C. {𝑎_𝑛 }_𝑛≥11 is bounded above by \(\frac{17}{8}\)
• D. \(∑^∞ _ {n=1} a_n\) is convergent

Answer 1

The Correct Option is B

To determine which statement about the sequence \( \{a_n\}_{n \geq 1} \) is true, we start by analyzing the recurrence relation given:

\[ 4a_{n+1} = \sqrt{45 + 16a_n} \]  

This relation can be rewritten as:

\[ a_{n+1} = \frac{\sqrt{45 + 16a_n}}{4} \]

Let's analyze the behavior of the sequence starting from \( a_1 = 1 \).

1. Substitute \( a_1 = 1 \) into the recurrence relation: \[ a_2 = \frac{\sqrt{45 + 16 \cdot 1}}{4} = \frac{\sqrt{61}}{4} \]
2. Observe if the sequence is increasing by comparing \( \frac{\sqrt{61}}{4} \) and 1: \[ \sqrt{61} \approx 7.81, \quad \frac{7.81}{4} \approx 1.9525 \] Since \( 1.9525 > 1 \), \( a_2 > a_1 \), indicating that \( \{a_n\} \) is increasing.

We need to check if the sequence converges and if so, identify the limit. Assume \( \lim_{n \to \infty} a_n = L \). Then, substituting \( a_n \) with \( L \) in the recurrence relation, we get:

\[ L = \frac{\sqrt{45 + 16L}}{4} \]

Squaring both sides:

\[ 16L^2 = 45 + 16L \]

Rearranging the equation, we have:

\[ 16L^2 - 16L - 45 = 0 \]

This is a quadratic equation in \( L \). Solving it using the quadratic formula:

\[ L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad \text{where } a = 16, b = -16, c = -45 \]

\(L = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \cdot 16 \cdot (-45)}}{2 \cdot 16}\)

\(L = \frac{16 \pm \sqrt{256 + 2880}}{32}\)

\(L = \frac{16 \pm \sqrt{3136}}{32}\)

\(L = \frac{16 \pm 56}{32}\)

1. Calculating the possible values for \( L \):
   • \(L_1 = \frac{72}{32} = \frac{9}{4}\)
   • \(L_2 = \frac{-40}{32} = -\frac{5}{4}\) (not possible as the sequence elements are positive)

Thus, the sequence is monotonically increasing and converges to \( \frac{9}{4} \).

Therefore, the correct statement is: \( \{a_n\}_{n \geq 1} \) is monotonically increasing and converges to \( \frac{9}{4} \).