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 sum of the first 10 terms is K, then `(4K)/(101)` is equal to

Explanation

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 will follow these steps:

### Step 1: Identify the pattern in the series
The series can be expressed in terms of sums of odd numbers and squares of sums of natural numbers. The \( n \)-th term can be represented as:
\[ T_n = \frac{1}{(1 + 3 + 5 + \ldots + (2n-1))(1 + 2 + 3 + \ldots + n)^2} \]

### Step 2: Calculate the sums in the denominator
The sum of the first \( n \) odd numbers is given by:
\[ 1 + 3 + 5 + \ldots + (2n-1) = n^2 \]
The sum of the first \( n \) natural numbers is given by:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2} \]
Thus, the square of this sum is:
\[ \left( \frac{n(n+1)}{2} \right)^2 = \frac{n^2(n+1)^2}{4} \]

### Step 3: Substitute back into the term
Now substituting these results back into \( T_n \):
\[ T_n = \frac{1}{n^2 \cdot \frac{n^2(n+1)^2}{4}} = \frac{4}{n^4(n+1)^2} \]

### Step 4: Write the series in terms of \( T_n \)
The series \( S \) can now be expressed as:
\[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{4}{n^4(n+1)^2} \]

### Step 5: Calculate the first 10 terms
To find \( K \), the sum of the first 10 terms, we compute:
\[ K = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} \frac{4}{n^4(n+1)^2} \]

Calculating each term:
- For \( n = 1 \): \( T_1 = \frac{4}{1^4 \cdot 2^2} = 1 \)
- For \( n = 2 \): \( T_2 = \frac{4}{2^4 \cdot 3^2} = \frac{4}{16 \cdot 9} = \frac{1}{36} \)
- For \( n = 3 \): \( T_3 = \frac{4}{3^4 \cdot 4^2} = \frac{4}{81 \cdot 16} = \frac{1}{324} \)
- For \( n = 4 \): \( T_4 = \frac{4}{4^4 \cdot 5^2} = \frac{4}{256 \cdot 25} = \frac{1}{1600} \)
- For \( n = 5 \): \( T_5 = \frac{4}{5^4 \cdot 6^2} = \frac{4}{625 \cdot 36} = \frac{1}{5625} \)
- For \( n = 6 \): \( T_6 = \frac{4}{6^4 \cdot 7^2} = \frac{4}{1296 \cdot 49} = \frac{1}{63504} \)
- For \( n = 7 \): \( T_7 = \frac{4}{7^4 \cdot 8^2} = \frac{4}{2401 \cdot 64} = \frac{1}{153664} \)
- For \( n = 8 \): \( T_8 = \frac{4}{8^4 \cdot 9^2} = \frac{4}{4096 \cdot 81} = \frac{1}{331776} \)
- For \( n = 9 \): \( T_9 = \frac{4}{9^4 \cdot 10^2} = \frac{4}{6561 \cdot 100} = \frac{1}{2624400} \)
- For \( n = 10 \): \( T_{10} = \frac{4}{10^4 \cdot 11^2} = \frac{4}{10000 \cdot 121} = \frac{1}{3025000} \)

### Step 6: Sum the first 10 terms
Calculating \( K \):
\[ K = 1 + \frac{1}{36} + \frac{1}{324} + \frac{1}{1600} + \frac{1}{5625} + \frac{1}{63504} + \frac{1}{153664} + \frac{1}{331776} + \frac{1}{2624400} + \frac{1}{3025000} \]
This sum can be approximated numerically.

### Step 7: Calculate \( \frac{4K}{101} \)
Finally, we compute:
\[ \frac{4K}{101} \]
Using the approximate value of \( K \) calculated from the sum of the first 10 terms.