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7. Let [tex]$\left\{a_n\right\}$[/tex] be a sequence of positive real numbers satisfying [tex]$\frac{4}{a_{n+1}}=\frac{3}{a_n}+\frac{a_n^3}{81}, n \geq 1, a_1=1$[/tex]. Then all the terms of the sequence lie in:

(a) [tex]$\left[\frac{1}{2}, \frac{3}{2}\right]$[/tex]
(b) [tex]$[0,1]$[/tex]
(c) [tex]$[1,2]$[/tex]
(d) [tex]$[1,3]$[/tex]



Answer :

GinnyAnswer
Given the sequence [tex]\( \{a_n\} \)[/tex] of positive real numbers defined by the recurrence relation:
[tex]\[ \frac{4}{a_{n+1}} = \frac{3}{a_n} + \frac{a_n^3}{81} \][/tex]
with the initial condition [tex]\( a_1 = 1 \)[/tex], we want to determine the range in which all terms of this sequence lie.

Let's compute the first few terms of the sequence to observe any pattern or range that the sequence falls into.

1. For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 1 \][/tex]

2. For [tex]\( n = 2 \)[/tex]:
[tex]\[ \frac{4}{a_2} = \frac{3}{a_1} + \frac{a_1^3}{81} \][/tex]
Substitute [tex]\( a_1 = 1 \)[/tex]:
[tex]\[ \frac{4}{a_2} = \frac{3}{1} + \frac{1^3}{81} \][/tex]
[tex]\[ \frac{4}{a_2} = 3 + \frac{1}{81} \][/tex]
[tex]\[ \frac{4}{a_2} = 3.012345679 \][/tex]
[tex]\[ a_2 = \frac{4}{3.012345679} \approx 1.3278688524590163 \][/tex]

3. For [tex]\( n = 3 \)[/tex]:
[tex]\[ \frac{4}{a_3} = \frac{3}{a_2} + \frac{a_2^3}{81} \][/tex]
Substitute [tex]\( a_2 \approx 1.3278688524590163 \)[/tex]:
[tex]\[ \frac{4}{a_3} = \frac{3}{1.3278688524590163} + \frac{(1.3278688524590163)^3}{81} \][/tex]
[tex]\[ \frac{4}{a_3} \approx 2.259 \][/tex]
[tex]\[ a_3 \approx 1.748125863047162 \][/tex]

4. For [tex]\( n = 4 \)[/tex]:
[tex]\[ \frac{4}{a_4} = \frac{3}{a_3} + \frac{a_3^3}{81} \][/tex]
Substitute [tex]\( a_3 \approx 1.748125863047162 \)[/tex]:
[tex]\[ a_4 \approx 2.244572746025354 \][/tex]

5. For [tex]\( n = 5 \)[/tex]:
[tex]\[ \frac{4}{a_5} = \frac{3}{a_4} + \frac{a_4^3}{81} \][/tex]
Substitute [tex]\( a_4 \approx 2.244572746025354 \)[/tex]:
[tex]\[ a_5 \approx 2.709720323930604 \][/tex]

6. For [tex]\( n = 6 \)[/tex]:
[tex]\[ \frac{4}{a_6} = \frac{3}{a_5} + \frac{a_5^3}{81} \][/tex]
Substitute [tex]\( a_5 \approx 2.709720323930604 \)[/tex]:
[tex]\[ a_6 \approx 2.956919295078352 \][/tex]

7. For [tex]\( n = 7 \)[/tex]:
[tex]\[ a_7 \approx 2.999067777153062 \][/tex]

8. For [tex]\( n = 8 \)[/tex]:
[tex]\[ a_8 \approx 2.999999565435316 \][/tex]

9. For [tex]\( n = 9 \)[/tex]:
[tex]\[ a_9 \approx 2.999999999999906 \][/tex]

10. For [tex]\( n = 10 \)[/tex]:
[tex]\[ a_{10} = 3.0 \][/tex]

From the calculated terms, we can observe that the terms start from 1 and get closer to 3 as [tex]\( n \)[/tex] increases.

Based on these observations, all the terms of the sequence lie within the interval [tex]\([1, 3]\)[/tex].

Therefore, the correct answer is:
(d) [tex]\([1, 3]\)[/tex]

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