The sum of n terms of the series :
5 + 11 + 19+ 29+ 41 . . . is

Options

A. `(n(n-1)(n-+7))/(5)`
B. `(n (n+2)(n+4))/(3)`
C. `(n(n+5)(n+1))/(9)`
D. `((n+1)(n+3)(n+7))/(5)`

Answer

To find the sum of the first n terms of the series: 5 + 11 + 19 + 29 + 41 + ..., we first need to identify the pattern in the series.

### Step 1: Identify the Pattern
The given series is:
- 5, 11, 19, 29, 41

Let's find the differences between consecutive terms:
- 11 - 5 = 6
- 19 - 11 = 8
- 29 - 19 = 10
- 41 - 29 = 12

The differences are: 6, 8, 10, 12, which are increasing by 2 each time. This indicates that the series is not an arithmetic progression (AP) but rather a quadratic sequence.

### Step 2: Formulate the nth Term
Since the differences are increasing linearly, we can express the nth term of the series as a quadratic function:
\[ a_n = An^2 + Bn + C \]

To find A, B, and C, we can use the first few terms:
- For n = 1: \( a_1 = 5 \)
- For n = 2: \( a_2 = 11 \)
- For n = 3: \( a_3 = 19 \)

This gives us the following equations:
1. \( A(1^2) + B(1) + C = 5 \) → \( A + B + C = 5 \) (Equation 1)
2. \( A(2^2) + B(2) + C = 11 \) → \( 4A + 2B + C = 11 \) (Equation 2)
3. \( A(3^2) + B(3) + C = 19 \) → \( 9A + 3B + C = 19 \) (Equation 3)

### Step 3: Solve the System of Equations
Now we will solve these equations step by step.

**Subtract Equation 1 from Equation 2:**
\[ (4A + 2B + C) - (A + B + C) = 11 - 5 \]
\[ 3A + B = 6 \] (Equation 4)

**Subtract Equation 2 from Equation 3:**
\[ (9A + 3B + C) - (4A + 2B + C) = 19 - 11 \]
\[ 5A + B = 8 \] (Equation 5)

**Now subtract Equation 4 from Equation 5:**
\[ (5A + B) - (3A + B) = 8 - 6 \]
\[ 2A = 2 \]
\[ A = 1 \]

**Substitute A back into Equation 4:**
\[ 3(1) + B = 6 \]
\[ B = 3 \]

**Substitute A and B back into Equation 1:**
\[ 1 + 3 + C = 5 \]
\[ C = 1 \]

### Step 4: Write the nth Term
Now we have:
\[ a_n = n^2 + 3n + 1 \]

### Step 5: Find the Sum of the First n Terms
The sum of the first n terms \( S_n \) can be calculated as:
\[ S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (k^2 + 3k + 1) \]

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

We can compute:
\[ S_n = \sum_{k=1}^{n} k^2 + 3\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]
\[ S_n = \frac{n(n+1)(2n+1)}{6} + 3 \cdot \frac{n(n+1)}{2} + n \]

### Step 6: Simplify the Expression
Combining these terms:
\[ S_n = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2} + n \]
\[ S_n = \frac{n(n+1)(2n+1)}{6} + \frac{9n(n+1)}{6} + \frac{6n}{6} \]
\[ S_n = \frac{n(n+1)(2n+1 + 9) + 6n}{6} \]
\[ S_n = \frac{n(n+1)(2n + 10) + 6n}{6} \]

### Final Result
Thus, we can express the sum of the first n terms as:
\[ S_n = \frac{n(n+1)(n+5)}{3} \]