For the series `S=1+(1)/((1+3))(1+2)^(2)+(1)/((1+3+5))(1+2+3)^(2)+(1)/((1+3+5+7))(1+2+3+4)^(2)+……………..,` if the `7^("th")` term is K, then `(K)/(4)` is equal to

Answer

To solve the given series \( S = 1 + \frac{1}{(1+3)(1+2)^2} + \frac{1}{(1+3+5)(1+2+3)^2} + \frac{1}{(1+3+5+7)(1+2+3+4)^2} + \ldots \), we need to find the 7th term \( K \) and then compute \( \frac{K}{4} \).

### Step-by-Step Solution:

1. **Identify the General Term**:
The \( r \)-th term can be expressed as:
\[ T_r = \frac{1}{(1 + 3 + 5 + \ldots + (2r - 1))(1 + 2 + 3 + \ldots + r)^2} \]

2. **Sum of Odd Numbers**:
The sum of the first \( r \) odd numbers is given by:
\[ 1 + 3 + 5 + \ldots + (2r - 1) = r^2 \]
Therefore, the denominator becomes \( r^2 \).

3. **Sum of Natural Numbers**:
The sum of the first \( r \) natural numbers is:
\[ 1 + 2 + 3 + \ldots + r = \frac{r(r + 1)}{2} \]
Thus, the square of this sum is:
\[ \left( \frac{r(r + 1)}{2} \right)^2 = \frac{r^2(r + 1)^2}{4} \]

4. **Substituting into the General Term**:
Now substituting these results into the expression for \( T_r \):
\[ T_r = \frac{1}{r^2 \cdot \frac{r^2(r + 1)^2}{4}} = \frac{4}{r^4(r + 1)^2} \]

5. **Finding the 7th Term**:
To find the 7th term \( K \), set \( r = 7 \):
\[ T_7 = \frac{4}{7^4(7 + 1)^2} = \frac{4}{7^4 \cdot 8^2} \]

6. **Calculating \( 7^4 \) and \( 8^2 \)**:
- \( 7^4 = 2401 \)
- \( 8^2 = 64 \)

Therefore:
\[ T_7 = \frac{4}{2401 \cdot 64} \]

7. **Calculating \( K \)**:
\[ K = \frac{4}{2401 \cdot 64} \]

8. **Finding \( \frac{K}{4} \)**:
\[ \frac{K}{4} = \frac{1}{2401 \cdot 64} \]

### Final Calculation:
Now we compute \( 2401 \cdot 64 \):
\[ 2401 \cdot 64 = 153664 \]
Thus, we have:
\[ \frac{K}{4} = \frac{1}{153664} \]

### Conclusion:
The final answer is:
\[ \frac{K}{4} = \frac{1}{153664} \]