Sum to n terms of the series `(1)/(1.4.7) + (1)/(4.7.10) + (1)/(7.10.13) +…….`

Answer

To find the sum to n terms of the series

\[ S_n = \frac{1}{1 \cdot 4 \cdot 7} + \frac{1}{4 \cdot 7 \cdot 10} + \frac{1}{7 \cdot 10 \cdot 13} + \ldots \]

we start by determining the general term of the series.

### Step 1: Identify the nth term
The nth term of the series can be expressed as:

\[ T_n = \frac{1}{(3n - 2)(3n + 1)(3n + 4)} \]

To verify this, we can substitute n = 1, 2, and 3:

- For n = 1: \( T_1 = \frac{1}{1 \cdot 4 \cdot 7} \)
- For n = 2: \( T_2 = \frac{1}{4 \cdot 7 \cdot 10} \)
- For n = 3: \( T_3 = \frac{1}{7 \cdot 10 \cdot 13} \)

This confirms that our nth term is correct.

### Step 2: Rewrite the nth term
Next, we can rewrite the nth term using partial fractions:

\[ T_n = \frac{1}{6} \left( \frac{1}{3n - 2} - \frac{1}{3n + 4} \right) \]

### Step 3: Sum the series
Now, we need to find the sum of the first n terms:

\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{1}{6} \left( \frac{1}{3k - 2} - \frac{1}{3k + 4} \right) \]

This can be simplified as:

\[ S_n = \frac{1}{6} \left( \sum_{k=1}^{n} \frac{1}{3k - 2} - \sum_{k=1}^{n} \frac{1}{3k + 4} \right) \]

### Step 4: Evaluate the telescoping series
The sums are telescoping series. When we write out the first few terms, we can see that many terms will cancel:

\[ S_n = \frac{1}{6} \left( \left( \frac{1}{1} - \frac{1}{3n + 4} \right) \right) \]

### Step 5: Final result
Thus, the sum of the first n terms is:

\[ S_n = \frac{1}{6} \left( 1 - \frac{1}{3n + 4} \right) = \frac{1}{6} - \frac{1}{6(3n + 4)} \]

### Conclusion
The final expression for the sum of the first n terms of the series is:

\[ S_n = \frac{1}{6} - \frac{1}{6(3n + 4)} \]