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 problem, we need to find the 7th term of the series given by:

\[ S = 1 + \frac{1}{(1+3)(1+2)^2} + \frac{1}{(1+3+5)(1+2+3)^2} + \frac{1}{(1+3+5+7)(1+2+3+4)^2} + \ldots \]

Let's denote the nth term of the series as \( T_n \).

### Step 1: Identify the nth term \( T_n \)

The nth term can be expressed as:

\[ T_n = \frac{1}{(1 + 3 + 5 + \ldots + (2n-1))(1 + 2 + 3 + \ldots + n)^2} \]

### Step 2: Calculate the sum of the first n odd numbers

The sum of the first n odd numbers is given by:

\[ 1 + 3 + 5 + \ldots + (2n-1) = n^2 \]

### Step 3: Calculate the sum of the first n natural numbers

The sum of the first n natural numbers is given by:

\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]

### Step 4: Substitute these sums into the nth term

Now substituting these results into \( T_n \):

\[ T_n = \frac{1}{n^2 \left(\frac{n(n + 1)}{2}\right)^2} \]

### Step 5: Simplify the expression for \( T_n \)

This simplifies to:

\[ T_n = \frac{1}{n^2 \cdot \frac{n^2(n + 1)^2}{4}} = \frac{4}{n^4(n + 1)^2} \]

### Step 6: Find the 7th term \( T_7 \)

Now we can find \( T_7 \):

\[ T_7 = \frac{4}{7^4(7 + 1)^2} = \frac{4}{7^4 \cdot 8^2} \]

Calculating \( 7^4 \) and \( 8^2 \):

- \( 7^4 = 2401 \)
- \( 8^2 = 64 \)

Thus,

\[ T_7 = \frac{4}{2401 \cdot 64} = \frac{4}{153664} \]

### Step 7: Simplify \( T_7 \)

Now simplifying \( T_7 \):

\[ T_7 = \frac{1}{38416} \]

### Step 8: Find \( K \) and \( \frac{K}{4} \)

Since \( K = T_7 \), we have:

\[ K = \frac{1}{38416} \]

Now, we need to calculate \( \frac{K}{4} \):

\[ \frac{K}{4} = \frac{1}{4 \cdot 38416} = \frac{1}{153664} \]

### Final Answer

Thus, the value of \( \frac{K}{4} \) is:

\[ \frac{K}{4} = \frac{1}{153664} \]