https://file.aiquickdraw.com/mathbot/file/1729395078923ksed26tc.jpg

Let's break down and solve this problem step by step.

### Given:
You are provided with a sequence \( L_m \) defined as:

\[
L_m = \sqrt[3]{\frac{m^3 + m^2 - m - 1}{m}}
\]

for \( m \geq 2 \).

It is asked to find the smallest \( n \) such that the product:

\[
L_2 \cdot L_3 \cdot L_4 \cdot \dots \cdot L_n > 34
\]

Once the value of \( n \) is found, you are asked to evaluate:

\[
\sqrt[3]{\frac{n + 2}{4}} + 34
\]

### Step 1: Calculate the product of terms \( L_m \)
We need to compute the terms of \( L_m \) for \( m = 2, 3, 4, \dots \) and find the smallest \( n \) such that their product exceeds 34.

For each \( m \), the term \( L_m \) is given by:

\[
L_m = \sqrt[3]{\frac{m^3 + m^2 - m - 1}{m}}
\]

#### \( L_2 \):
\[
L_2 = \sqrt[3]{\frac{2^3 + 2^2 - 2 - 1}{2}} = \sqrt[3]{\frac{8 + 4 - 2 - 1}{2}} = \sqrt[3]{\frac{9}{2}} = \sqrt[3]{4.5}
\]
Approximating:
\[
L_2 \approx 1.65
\]

#### \( L_3 \):
\[
L_3 = \sqrt[3]{\frac{3^3 + 3^2 - 3 - 1}{3}} = \sqrt[3]{\frac{27 + 9 - 3 - 1}{3}} = \sqrt[3]{\frac{32}{3}} \approx \sqrt[3]{10.67}
\]
Approximating:
\[
L_3 \approx 2.17
\]

#### \( L_4 \):
\[
L_4 = \sqrt[3]{\frac{4^3 + 4^2 - 4 - 1}{4}} = \sqrt[3]{\frac{64 + 16 - 4 - 1}{4}} = \sqrt[3]{\frac{75}{4}} \approx \sqrt[3]{18.75}
\]
Approximating:
\[
L_4 \approx 2.65
\]

#### \( L_5 \):
\[
L_5 = \sqrt[3]{\frac{5^3 + 5^2 - 5 - 1}{5}} = \sqrt[3]{\frac{125 + 25 - 5 - 1}{5}} = \sqrt[3]{\frac{144}{5}} \approx \sqrt[3]{28.8}
\]
Approximating:
\[
L_5 \approx 3.07
\]

### Step 2: Compute the product \( L_2 \cdot L_3 \cdot L_4 \cdot \dots \)
We now compute the cumulative product until the product exceeds 34.

\[
L_2 \cdot L_3 = 1.65 \times 2.17 \approx 3.58
\]
\[
L_2 \cdot L_3 \cdot L_4 = 3.58 \times 2.65 \approx 9.49
\]
\[
L_2 \cdot L_3 \cdot L_4 \cdot L_5 = 9.49 \times 3.07 \approx 29.13
\]

Clearly, \( n = 5 \) does not satisfy the condition \( L_2 \cdot L_3 \cdot L_4 \cdot L_5 > 34 \).

#### \( L_6 \):
\[
L_6 = \sqrt[3]{\frac{6^3 + 6^2 - 6 - 1}{6}} = \sqrt[3]{\frac{216 + 36 - 6 - 1}{6}} = \sqrt[3]{\frac{245}{6}} \approx \sqrt[3]{40.83}
\]
Approximating:
\[
L_6 \approx 3.44
\]
Now, multiplying by \( L_6 \):

\[
L_2 \cdot L_3 \cdot L_4 \cdot L_5 \cdot L_6 = 29.13 \times 3.44 \approx 100.23
\]

Thus, \( n = 6 \) satisfies \( L_2 \cdot L_3 \cdot L_4 \cdot L_5 \cdot L_6 > 34 \).

### Step 3: Compute the final expression
Now that we know \( n = 6 \), we evaluate the final expression:

\[
\sqrt[3]{\frac{n + 2}{4}} + 34 = \sqrt[3]{\frac{6 + 2}{4}} + 34 = \sqrt[3]{\frac{8}{4}} + 34 = \sqrt[3]{2} + 34
\]

Approximating:

\[
\sqrt[3]{2} \approx 1.26
\]
Thus, the final value is:
\[
1.26 + 34 = 35.26
\]
Rounding gives approximately 35.

It appears that the closest option is **65**. However, based on the image, this might be due to a typo or rounding mismatch in the actual options.

Would you like more details on any part? Here are 5 additional questions related to this topic:

1. What is the cube root of other common fractions?
2. How is the cube root function used in real-world applications?
3. What happens to the product \( L_2 \cdot L_3 \cdot \dots \) as \( n \) increases indefinitely?
4. How can you calculate cube roots by hand more accurately?
5. What are the properties of the cubic functions in sequences?

Tip: Always round intermediate calculations carefully to avoid discrepancies in final answers.

Diberikan barisan (L_m) dengan definisi L_m = √[3]{m^3 + m^2 - m - 1 / m} untuk m ≥ 2 dan n adalah bilangan asli terkecil yang memenuhi L_2 · L_3 · L_4 · ... · L_n > 34. Nilai dari √[3]{n + 2 / 4} + 34 adalah ...

Learn how to solve a problem involving sequences defined by cubic roots and inequalities. The task involves finding the smallest integer n such that the product of specific terms exceeds a given value and calculating a final expression involving cube roots. Suitable for high school students in Grades 10-12.

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