`2 + 4 + 7 + 11 + 16 +`.... to `n` term =

Options

A. `(1)/(6) (n^(2) + 3n + 8)`
B. `(n)/(6) (n^(2) + 3n + 8)`
C. `(1)/(6) (n^(2) - 3n + 8)`
D. `(n)/(6) (n^(2) - 3n + 8)`

Answer

To find the sum of the sequence \(2 + 4 + 7 + 11 + 16 + \ldots\) up to \(n\) terms, we first need to identify the pattern in the sequence.

### Step 1: Identify the sequence
The given sequence is:
- First term \(T_1 = 2\)
- Second term \(T_2 = 4\)
- Third term \(T_3 = 7\)
- Fourth term \(T_4 = 11\)
- Fifth term \(T_5 = 16\)

Now, let's find the differences between consecutive terms:
- \(T_2 - T_1 = 4 - 2 = 2\)
- \(T_3 - T_2 = 7 - 4 = 3\)
- \(T_4 - T_3 = 11 - 7 = 4\)
- \(T_5 - T_4 = 16 - 11 = 5\)

The differences are \(2, 3, 4, 5\), which themselves increase by \(1\) each time. This indicates that the sequence is not an arithmetic progression but a quadratic sequence.

### Step 2: Formulate the \(n\)th term
Since the differences of the terms form an arithmetic sequence, we can express the \(n\)th term \(T_n\) in the form of a quadratic equation:
\[ T_n = an^2 + bn + c \]

### Step 3: Set up equations
Using the first three terms, we can set up the following equations:
1. For \(n=1\): \(a(1)^2 + b(1) + c = 2 \quad \Rightarrow \quad a + b + c = 2\) (Equation 1)
2. For \(n=2\): \(a(2)^2 + b(2) + c = 4 \quad \Rightarrow \quad 4a + 2b + c = 4\) (Equation 2)
3. For \(n=3\): \(a(3)^2 + b(3) + c = 7 \quad \Rightarrow \quad 9a + 3b + c = 7\) (Equation 3)

### Step 4: Solve the equations
Now we can solve these equations step by step.

**Subtract Equation 1 from Equation 2:**
\[ (4a + 2b + c) - (a + b + c) = 4 - 2 \]
\[ 3a + b = 2 \quad \text{(Equation 4)} \]

**Subtract Equation 2 from Equation 3:**
\[ (9a + 3b + c) - (4a + 2b + c) = 7 - 4 \]
\[ 5a + b = 3 \quad \text{(Equation 5)} \]

**Now subtract Equation 4 from Equation 5:**
\[ (5a + b) - (3a + b) = 3 - 2 \]
\[ 2a = 1 \quad \Rightarrow \quad a = \frac{1}{2} \]

**Substituting \(a\) back into Equation 4:**
\[ 3\left(\frac{1}{2}\right) + b = 2 \]
\[ \frac{3}{2} + b = 2 \quad \Rightarrow \quad b = 2 - \frac{3}{2} = \frac{1}{2} \]

**Substituting \(a\) and \(b\) back into Equation 1:**
\[ \frac{1}{2} + \frac{1}{2} + c = 2 \]
\[ 1 + c = 2 \quad \Rightarrow \quad c = 1 \]

### Step 5: Write the \(n\)th term
Now we have:
\[ T_n = \frac{1}{2}n^2 + \frac{1}{2}n + 1 \]

### Step 6: Find the sum of the first \(n\) terms
The sum \(S_n\) of the first \(n\) terms can be calculated as:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left(\frac{1}{2}k^2 + \frac{1}{2}k + 1\right) \]
This can be separated into three sums:
\[ S_n = \frac{1}{2} \sum_{k=1}^{n} k^2 + \frac{1}{2} \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]

Using the formulas for these sums:
- \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)
- \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)
- \(\sum_{k=1}^{n} 1 = n\)

Substituting these into the equation gives:
\[ S_n = \frac{1}{2} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{1}{2} \cdot \frac{n(n+1)}{2} + n \]

### Step 7: Simplify the sum
Combining these terms will yield the final expression for \(S_n\).