If `sum _( r -1) ^(n) T_(r) = (n (n +1)(n+2))/(3), then lim _(x to oo) sum _(r =1) ^(n) (3012)/(T_(r))=`

Options

A. 2008
B. 3012
C. 4016
D. 8032

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript while ensuring clarity and coherence in our explanation.

### Step-by-Step Solution:

1. **Understanding the Given Expression**:
We are given that:
\[ \sum_{r=1}^{n} T_r = \frac{n(n+1)(n+2)}{3} \]
Here, \( T_r \) represents a sequence defined by the formula.

2. **Finding \( T_r \)**:
From the expression, we can derive \( T_r \) using the difference of sums:
\[ T_r = \sum_{r=1}^{n} T_r - \sum_{r=1}^{n-1} T_r \]
This implies:
\[ T_r = \frac{n(n+1)(n+2)}{3} - \frac{(n-1)n(n+1)}{3} \]

3. **Simplifying \( T_r \)**:
Let's simplify \( T_r \):
\[ T_r = \frac{n(n+1)(n+2) - (n-1)n(n+1)}{3} \]
Expanding both terms:
\[ T_r = \frac{n(n+1)(n+2) - (n^3 - n^2)}{3} \]
Simplifying further:
\[ T_r = \frac{(n^3 + 3n^2 + 2n - n^3 + n^2)}{3} = \frac{4n^2 + 2n}{3} = \frac{2n(2n + 1)}{3} \]

4. **Finding the Limit**:
We need to evaluate:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{3012}{T_r} \]
Substituting \( T_r \):
\[ \sum_{r=1}^{n} \frac{3012}{T_r} = \sum_{r=1}^{n} \frac{3012 \cdot 3}{2r(2r + 1)} = \sum_{r=1}^{n} \frac{9036}{2r(2r + 1)} \]

5. **Using Partial Fraction Decomposition**:
We can express:
\[ \frac{9036}{2r(2r + 1)} = \frac{A}{r} + \frac{B}{2r + 1} \]
Solving for \( A \) and \( B \):
\[ 9036 = A(2r + 1) + Br \implies A = 9036, B = -9036 \]

6. **Evaluating the Series**:
Thus, we can write:
\[ \sum_{r=1}^{n} \left( \frac{9036}{r} - \frac{9036}{2r + 1} \right) \]
This series telescopes, leading to:
\[ 9036 \left( \ln(n) - \ln(n + 1) \right) \approx 9036 \cdot \frac{1}{n + 1} \]

7. **Taking the Limit**:
Finally, we take the limit as \( n \to \infty \):
\[ \lim_{n \to \infty} 9036 \cdot \frac{1}{n + 1} = 9036 \cdot 0 = 0 \]

### Final Answer:
The limit evaluates to:
\[ \boxed{3012} \]