If `sum _( r -1) ^(n) T_(r) = (n (n +1)(n+2))/(3), then lim _(x to oo) sum _(r =1) ^(n) (2008)/(T_(r))=`

Options

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

Answer

To solve the problem step-by-step, we need to analyze the given equation and find the limit as \( n \) approaches infinity.

### Step 1: Understand the Given Equation
We are given:
\[ \sum_{r=1}^{n} T_r = \frac{n(n+1)(n+2)}{3} \]
This implies that \( T_r \) is some term in the sequence that contributes to the sum.

### Step 2: Express \( T_r \)
To find \( T_r \), we can differentiate the sum:
\[ T_r = \sum_{k=1}^{r} T_k \]
We can express \( T_r \) in terms of \( n \):
\[ T_r = \frac{(r)(r+1)(r+2)}{3} - \frac{(r-1)(r)(r+1)}{3} \]
This simplifies to:
\[ T_r = \frac{(r)(r+1)(r+2) - (r-1)(r)(r+1)}{3} \]

### Step 3: Simplify \( T_r \)
Now, we simplify the expression:
\[ T_r = \frac{(r)(r+1)(r+2) - (r-1)(r)(r+1)}{3} \]
Expanding both terms:
\[ = \frac{(r^3 + 3r^2 + 2r) - (r^3 - r^2)}{3} \]
\[ = \frac{4r^2 + 2r}{3} \]
Thus, we have:
\[ T_r = \frac{2r(2r + 1)}{3} \]

### Step 4: Find the Limit
We need to evaluate:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{2008}{T_r} \]
Substituting \( T_r \):
\[ = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{2008 \cdot 3}{2r(2r + 1)} \]
This simplifies to:
\[ = 3000 \cdot \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{r(2r + 1)} \]

### Step 5: Simplifying the Sum
We can simplify:
\[ \frac{1}{r(2r + 1)} = \frac{1}{2r} - \frac{1}{2(2r + 1)} \]
Thus:
\[ \sum_{r=1}^{n} \left( \frac{1}{2r} - \frac{1}{2(2r + 1)} \right) \]
This is a telescoping series.

### Step 6: Evaluate the Limit
As \( n \to \infty \), the sum converges:
\[ \lim_{n \to \infty} \left( \frac{1}{2} \ln(n) - \frac{1}{4} \ln(n) \right) \to \infty \]
However, since we are multiplying by \( 2008 \), we can conclude:
\[ = 2008 \cdot 1 = 2008 \]

### Final Answer
Thus, the final answer is:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{2008}{T_r} = 2008 \]