Find the sum of the series ` 3 + 7 + 13 + 21 + 31 + …`
to n terms . Find the sun when` n=20`

Options

A. `(n(n^(2) + 2n + 3))/(3), 3110`
B. `(n(n^(2) + 3n + 5))/(3), 3110`
C. `(n(n^(2) + 3n + 5))/(3) , 3120`
D. `(n(n^(2) + 2n +2))/(3), 3120`

Answer

To find the sum of the series \(3 + 7 + 13 + 21 + 31 + \ldots\) up to \(n\) terms, we will first analyze the pattern in the series.

### Step 1: Identify the pattern in the series
The given series is:
- First term: \(3\)
- Second term: \(7\)
- Third term: \(13\)
- Fourth term: \(21\)
- Fifth term: \(31\)

Let's calculate the differences between consecutive terms:
- \(7 - 3 = 4\)
- \(13 - 7 = 6\)
- \(21 - 13 = 8\)
- \(31 - 21 = 10\)

The differences are \(4, 6, 8, 10\), which are increasing by \(2\). This indicates that the series can be expressed as a quadratic sequence.

### Step 2: Formulate the \(n\)th term
The \(n\)th term can be expressed as:
\[ T_n = an^2 + bn + c \]
We need to find the coefficients \(a\), \(b\), and \(c\).

Using the first three terms:
1. For \(n = 1\): \(T_1 = 3\) gives \(a(1)^2 + b(1) + c = 3\) → \(a + b + c = 3\)
2. For \(n = 2\): \(T_2 = 7\) gives \(a(2)^2 + b(2) + c = 7\) → \(4a + 2b + c = 7\)
3. For \(n = 3\): \(T_3 = 13\) gives \(a(3)^2 + b(3) + c = 13\) → \(9a + 3b + c = 13\)

We now have a system of equations:
1. \(a + b + c = 3\) (1)
2. \(4a + 2b + c = 7\) (2)
3. \(9a + 3b + c = 13\) (3)

### Step 3: Solve the system of equations
Subtract equation (1) from (2):
\[ (4a + 2b + c) - (a + b + c) = 7 - 3 \]
This simplifies to:
\[ 3a + b = 4 \quad (4) \]

Subtract equation (2) from (3):
\[ (9a + 3b + c) - (4a + 2b + c) = 13 - 7 \]
This simplifies to:
\[ 5a + b = 6 \quad (5) \]

Now subtract equation (4) from (5):
\[ (5a + b) - (3a + b) = 6 - 4 \]
This simplifies to:
\[ 2a = 2 \quad \Rightarrow \quad a = 1 \]

Substituting \(a = 1\) into equation (4):
\[ 3(1) + b = 4 \quad \Rightarrow \quad b = 1 \]

Substituting \(a = 1\) and \(b = 1\) into equation (1):
\[ 1 + 1 + c = 3 \quad \Rightarrow \quad c = 1 \]

Thus, the \(n\)th term is:
\[ T_n = n^2 + n + 1 \]

### Step 4: Find the sum of the series up to \(n\) terms
The sum \(S_n\) of the first \(n\) terms is given by:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + k + 1) \]
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 for the sums:
- \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)
- \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)
- \(\sum_{k=1}^{n} 1 = n\)

Thus,
\[ S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + n \]

### Step 5: Simplify the expression for \(S_n\)
Combine the terms:
\[ S_n = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6} + \frac{6n}{6} \]
\[ S_n = \frac{n(n+1)(2n+1 + 3) + 6n}{6} \]
\[ S_n = \frac{n(n+1)(2n + 4) + 6n}{6} \]
\[ S_n = \frac{n(n+1)(2n + 4) + 6n}{6} \]

### Step 6: Calculate \(S_{20}\)
Now, substituting \(n = 20\):
\[ S_{20} = \frac{20(21)(2(20) + 4) + 6(20)}{6} \]
Calculating:
\[ = \frac{20(21)(40 + 4) + 120}{6} \]
\[ = \frac{20(21)(44) + 120}{6} \]
\[ = \frac{18480 + 120}{6} \]
\[ = \frac{18600}{6} = 3100 \]

### Final Answer
The sum of the series \(3 + 7 + 13 + 21 + 31 + \ldots\) up to \(20\) terms is \(3100\).