If `sum _(r =1) ^(n ) T _(r) = (n +1) ( n +2) ( n +3)` then find ` sum _( r =1) ^(n) (1)/(T _(r))`

Answer

To solve the problem, we need to find the summation \( \sum_{r=1}^{n} \frac{1}{T_r} \) given that \( \sum_{r=1}^{n} T_r = (n + 1)(n + 2)(n + 3) \).

### Step-by-Step Solution:

1. **Define the Summation**:
We have:
\[ S_n = \sum_{r=1}^{n} T_r = (n + 1)(n + 2)(n + 3) \]

2. **Find \( S_{n-1} \)**:
We need to find \( S_{n-1} \):
\[ S_{n-1} = (n)(n + 1)(n + 2) \]

3. **Determine \( T_n \)**:
Using the relationship \( T_n = S_n - S_{n-1} \):
\[ T_n = (n + 1)(n + 2)(n + 3) - (n)(n + 1)(n + 2) \]
Factor out \( (n + 1)(n + 2) \):
\[ T_n = (n + 1)(n + 2) \left[ (n + 3) - n \right] = (n + 1)(n + 2)(3) \]
Thus, we have:
\[ T_n = 3(n + 1)(n + 2) \]

4. **Find \( \frac{1}{T_r} \)**:
Now, we can express \( \frac{1}{T_r} \):
\[ \frac{1}{T_r} = \frac{1}{3(r + 1)(r + 2)} \]

5. **Set Up the Summation**:
We need to find:
\[ \sum_{r=1}^{n} \frac{1}{T_r} = \sum_{r=1}^{n} \frac{1}{3(r + 1)(r + 2)} = \frac{1}{3} \sum_{r=1}^{n} \frac{1}{(r + 1)(r + 2)} \]

6. **Partial Fraction Decomposition**:
We can decompose \( \frac{1}{(r + 1)(r + 2)} \):
\[ \frac{1}{(r + 1)(r + 2)} = \frac{1}{r + 1} - \frac{1}{r + 2} \]

7. **Evaluate the Summation**:
Now substituting back:
\[ \sum_{r=1}^{n} \left( \frac{1}{r + 1} - \frac{1}{r + 2} \right) \]
This is a telescoping series:
\[ = \left( \frac{1}{2} - \frac{1}{n + 2} \right) \]

8. **Final Expression**:
Therefore, we have:
\[ \sum_{r=1}^{n} \frac{1}{T_r} = \frac{1}{3} \left( \frac{1}{2} - \frac{1}{n + 2} \right) \]
Simplifying gives:
\[ = \frac{1}{6} - \frac{1}{3(n + 2)} \]

### Final Answer:
\[ \sum_{r=1}^{n} \frac{1}{T_r} = \frac{n}{6(n + 2)} \]