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 value of 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. **Understanding the Given Summation**:
We know that:
\[ \sum_{r=1}^{n} T_r = (n+1)(n+2)(n+3) \]
This implies that \( T_r \) is a sequence whose sum can be expressed in this polynomial form.

2. **Finding the General Term \( T_r \)**:
To find \( T_r \), we can express it in terms of \( n \):
\[ T_r = (n+1)(n+2)(n+3) - \sum_{k=1}^{r-1} T_k \]
Since we are looking for a specific form of \( T_r \), we can differentiate the polynomial to find a pattern.

3. **Using the Polynomial Expansion**:
The expression \( (n+1)(n+2)(n+3) \) expands to:
\[ n^3 + 6n^2 + 11n + 6 \]
To find \( T_r \), we can derive the differences between successive terms.

4. **Finding \( T_r \)**:
The \( r \)-th term can be derived as:
\[ T_r = (r+1)(r+2)(r+3) - (r)(r+1)(r+2) \]
This gives us a specific polynomial for each \( r \).

5. **Finding the Summation**:
We need to find:
\[ \sum_{r=1}^{n} \frac{1}{T_r} \]
From the previous steps, we can express \( \frac{1}{T_r} \) as:
\[ \frac{1}{T_r} = \frac{1}{3} \left( \frac{1}{n+1} - \frac{1}{n+2} \right) \]

6. **Evaluating the Summation**:
Now substituting this back into the summation:
\[ \sum_{r=1}^{n} \frac{1}{T_r} = \frac{1}{3} \sum_{r=1}^{n} \left( \frac{1}{n+1} - \frac{1}{n+2} \right) \]
This telescopes to:
\[ = \frac{1}{3} \left( \frac{1}{2} - \frac{1}{n+2} \right) \]

7. **Final Calculation**:
Taking the common denominator:
\[ = \frac{1}{3} \left( \frac{n+2 - 2}{2(n+2)} \right) = \frac{1}{3} \cdot \frac{n}{2(n+2)} = \frac{n}{6(n+2)} \]

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