If `Sigma_(r=1)^(n)t_(r)=(1)/(6)n(n+1)(n+2), AA n ge 1,` then the value of `lim_(nrarroo)Sigma_(r=1)^(n)(1)/(t_(r))` is equal to

Options

A. 2
B. 3
C. `(3)/(2)`
D. 6

Answer

To solve the problem, we need to find the limit:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{t_r} \]

where it is given that:

\[ \sum_{r=1}^{n} t_r = \frac{1}{6} n(n+1)(n+2) \]

### Step 1: Find the expression for \( t_n \)

From the definition of \( t_n \):

\[ t_n = \sum_{r=1}^{n} t_r - \sum_{r=1}^{n-1} t_r \]

Using the given formula for \( \sum_{r=1}^{n} t_r \):

\[ \sum_{r=1}^{n-1} t_r = \frac{1}{6} (n-1)n(n+1) \]

Now, substituting these into the equation for \( t_n \):

\[ t_n = \frac{1}{6} n(n+1)(n+2) - \frac{1}{6} (n-1)n(n+1) \]

### Step 2: Simplify \( t_n \)

Now we simplify \( t_n \):

\[ t_n = \frac{1}{6} \left[ n(n+1)(n+2) - (n-1)n(n+1) \right] \]

Expanding both terms:

1. \( n(n+1)(n+2) = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n \)
2. \( (n-1)n(n+1) = n(n^2 + n - n - 1) = n^3 - n^2 \)

Subtracting these:

\[ t_n = \frac{1}{6} \left[ (n^3 + 3n^2 + 2n) - (n^3 - n^2) \right] \]

This simplifies to:

\[ t_n = \frac{1}{6} \left[ 4n^2 + 2n \right] = \frac{2n^2 + n}{3} \]

### Step 3: Find \( \frac{1}{t_r} \)

Now we find \( \frac{1}{t_r} \):

\[ \frac{1}{t_r} = \frac{3}{2r^2 + r} \]

### Step 4: Find the sum \( \sum_{r=1}^{n} \frac{1}{t_r} \)

We can rewrite the sum:

\[ \sum_{r=1}^{n} \frac{1}{t_r} = \sum_{r=1}^{n} \frac{3}{2r^2 + r} \]

### Step 5: Analyze the limit

As \( n \to \infty \), we can approximate:

\[ \frac{1}{t_r} \approx \frac{3}{2r^2} \]

Thus,

\[ \sum_{r=1}^{n} \frac{1}{t_r} \approx \sum_{r=1}^{n} \frac{3}{2r^2} \]

This sum can be approximated by the integral:

\[ \sum_{r=1}^{n} \frac{1}{r^2} \approx \int_{1}^{n} \frac{1}{x^2} \, dx = 1 - \frac{1}{n} \to 1 \text{ as } n \to \infty \]

Hence,

\[ \sum_{r=1}^{n} \frac{3}{2r^2} \approx \frac{3}{2} \cdot 1 = \frac{3}{2} \]

### Step 6: Final limit

Thus, the limit becomes:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{t_r} = \frac{3}{2} \]

### Conclusion

The final answer is:

\[ \boxed{3} \]