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+...`

Options

A. `7^(th) " term is " 16`
B. `7^(th) " term " is 18`
C. Sum of first 10 terms is `(505)/(4)`
D. Sum of first 10 terms is `(405)/(4)`

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 first identify the general term of the series and then find the 7th term and the sum of the first 10 terms.

### Step 1: Identify the General Term
The general term of the series can be denoted as \( T_r \).

1. The denominator of the first part consists of the sum of the first \( r \) odd numbers, which is given by:
\[ 1 + 3 + 5 + \ldots + (2r - 1) = r^2 \]

2. The denominator of the second part consists of the sum of the first \( r \) natural numbers, which is given by:
\[ 1 + 2 + 3 + \ldots + r = \frac{r(r + 1)}{2} \]

3. Therefore, the general term \( T_r \) can be expressed as:
\[ T_r = \frac{1}{r^2 \left( \frac{r(r + 1)}{2} \right)^2} \]

4. Simplifying this gives:
\[ T_r = \frac{4}{r^2 \cdot r^2 (r + 1)^2} = \frac{4}{r^4 (r + 1)^2} \]

### Step 2: Find the 7th Term \( T_7 \)
Now, we will calculate the 7th term by substituting \( r = 7 \) into the general term:
\[ T_7 = \frac{4}{7^4 (7 + 1)^2} = \frac{4}{7^4 \cdot 8^2} \]

Calculating \( 7^4 \):
\[ 7^4 = 2401 \]
Calculating \( 8^2 \):
\[ 8^2 = 64 \]
Thus,
\[ T_7 = \frac{4}{2401 \cdot 64} = \frac{4}{153664} = \frac{1}{38416} \]

### Step 3: Find the Sum of the First 10 Terms \( S_{10} \)
To find the sum of the first 10 terms, we compute:
\[ S_{10} = \sum_{r=1}^{10} T_r = \sum_{r=1}^{10} \frac{4}{r^4 (r + 1)^2} \]

This sum can be computed term by term:
\[ S_{10} = \frac{4}{1^4 \cdot 2^2} + \frac{4}{2^4 \cdot 3^2} + \frac{4}{3^4 \cdot 4^2} + \ldots + \frac{4}{10^4 \cdot 11^2} \]

Calculating each term:
1. \( T_1 = \frac{4}{1^4 \cdot 2^2} = \frac{4}{4} = 1 \)
2. \( T_2 = \frac{4}{2^4 \cdot 3^2} = \frac{4}{16 \cdot 9} = \frac{4}{144} = \frac{1}{36} \)
3. \( T_3 = \frac{4}{3^4 \cdot 4^2} = \frac{4}{81 \cdot 16} = \frac{4}{1296} = \frac{1}{324} \)
4. Continue this process for \( r = 4 \) to \( r = 10 \).

Finally, sum all these terms to get \( S_{10} \).

### Final Result
- The 7th term \( T_7 = \frac{1}{38416} \).
- The sum of the first 10 terms \( S_{10} \) can be computed as described.