If `S_(n) = 1 + 3 + 7 + 13 + 21 + "….."` upto n terms, then

Options

A. `S_(10) = 340`
B. `T_(10) = 91`
C. `S_(10) = 430`
D. `T_(6) = 31`

Answer

To solve the problem, we need to find the sum \( S_n = 1 + 3 + 7 + 13 + 21 + \ldots \) up to \( n \) terms.

### Step-by-step Solution:

1. **Identify the Sequence**:
The given sequence is \( 1, 3, 7, 13, 21, \ldots \). We need to find the pattern in the differences between the terms.

- The first term is \( 1 \).
- The second term is \( 3 \) (difference of \( 2 \)).
- The third term is \( 7 \) (difference of \( 4 \)).
- The fourth term is \( 13 \) (difference of \( 6 \)).
- The fifth term is \( 21 \) (difference of \( 8 \)).

The differences are \( 2, 4, 6, 8, \ldots \), which form an arithmetic sequence with a common difference of \( 2 \).

2. **Find the General Term**:
The \( n \)-th term can be expressed in terms of \( n \). The differences suggest that the \( n \)-th term can be represented as:
\[ t_n = t_{n-1} + (2(n-1)) \]
Starting from \( t_1 = 1 \):
- \( t_2 = 1 + 2 = 3 \)
- \( t_3 = 3 + 4 = 7 \)
- \( t_4 = 7 + 6 = 13 \)
- \( t_5 = 13 + 8 = 21 \)

We can derive a formula for \( t_n \):
\[ t_n = 1 + 2(1 + 2 + \ldots + (n-1)) \]
The sum of the first \( n-1 \) natural numbers is \( \frac{(n-1)n}{2} \), thus:
\[ t_n = 1 + 2 \cdot \frac{(n-1)n}{2} = 1 + (n-1)n = n^2 - n + 1 \]

3. **Calculate the Sum \( S_n \)**:
Now we can find \( S_n \):
\[ S_n = \sum_{k=1}^{n} t_k = \sum_{k=1}^{n} (k^2 - k + 1) \]
This can be separated into three sums:
\[ S_n = \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]
Using the formulas:
- \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \)
- \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)
- \( \sum_{k=1}^{n} 1 = n \)

We can substitute these into the equation:
\[ S_n = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} + n \]

4. **Simplifying \( S_n \)**:
To simplify:
\[ S_n = \frac{n(n+1)(2n+1)}{6} - \frac{3n(n+1)}{6} + \frac{6n}{6} \]
\[ S_n = \frac{n(n+1)(2n+1 - 3) + 6n}{6} \]
\[ S_n = \frac{n(n+1)(2n - 2) + 6n}{6} \]
\[ S_n = \frac{n(n+1)(2(n-1)) + 6n}{6} \]
\[ S_n = \frac{n(n+1)(2n - 2) + 6n}{6} \]

5. **Final Calculation for \( S_{10} \)**:
Now let's calculate \( S_{10} \):
\[ S_{10} = \frac{10(11)(2(10) + 1)}{6} - \frac{10(11)}{2} + 10 \]
After substituting \( n = 10 \) and simplifying, we find:
\[ S_{10} = 340 \]

### Final Answer:
Thus, the sum \( S_n \) up to \( n \) terms is given by the derived formula, and specifically for \( n = 10 \), \( S_{10} = 340 \).