If the sum of first 15 terms of the series 3+7+14+24+37+... Is 15k, then k is equle to

Options

A. 126
B. 122
C. 81
D. 119

Answer

To solve the problem, we need to find the value of \( k \) such that the sum of the first 15 terms of the series \( 3, 7, 14, 24, 37, \ldots \) is equal to \( 15k \).

### Step-by-Step Solution:

1. **Identify the Series**:
The series given is \( 3, 7, 14, 24, 37, \ldots \). We need to find a pattern or formula for the \( n \)-th term of the series.

2. **Find the Differences**:
Let's find the differences between consecutive terms:
- \( 7 - 3 = 4 \)
- \( 14 - 7 = 7 \)
- \( 24 - 14 = 10 \)
- \( 37 - 24 = 13 \)

The first differences are \( 4, 7, 10, 13 \).

3. **Find the Second Differences**:
Now, let's find the differences of the first differences:
- \( 7 - 4 = 3 \)
- \( 10 - 7 = 3 \)
- \( 13 - 10 = 3 \)

The second differences are constant and equal to \( 3 \), indicating that the \( n \)-th term is a quadratic function.

4. **Assume a Quadratic Formula**:
We can assume the \( n \)-th term \( T_n \) can be expressed as:
\[ T_n = an^2 + bn + c \]

5. **Set Up Equations**:
Using the first three terms:
- For \( n = 1 \): \( T_1 = 3 \) → \( a(1)^2 + b(1) + c = 3 \) → \( a + b + c = 3 \)
- For \( n = 2 \): \( T_2 = 7 \) → \( a(2)^2 + b(2) + c = 7 \) → \( 4a + 2b + c = 7 \)
- For \( n = 3 \): \( T_3 = 14 \) → \( a(3)^2 + b(3) + c = 14 \) → \( 9a + 3b + c = 14 \)

6. **Solve the System of Equations**:
We have the following system:
\[ \begin{align*} 1. & \quad a + b + c = 3 \\ 2. & \quad 4a + 2b + c = 7 \\ 3. & \quad 9a + 3b + c = 14 \end{align*} \]

Subtract the first equation from the second:
\[ (4a + 2b + c) - (a + b + c) = 7 - 3 \implies 3a + b = 4 \quad \text{(Equation 4)} \]

Subtract the second equation from the third:
\[ (9a + 3b + c) - (4a + 2b + c) = 14 - 7 \implies 5a + b = 7 \quad \text{(Equation 5)} \]

Now, subtract Equation 4 from Equation 5:
\[ (5a + b) - (3a + b) = 7 - 4 \implies 2a = 3 \implies a = \frac{3}{2} \]

Substitute \( a \) back into Equation 4:
\[ 3\left(\frac{3}{2}\right) + b = 4 \implies \frac{9}{2} + b = 4 \implies b = 4 - \frac{9}{2} = \frac{8}{2} - \frac{9}{2} = -\frac{1}{2} \]

Substitute \( a \) and \( b \) back into Equation 1:
\[ \frac{3}{2} - \frac{1}{2} + c = 3 \implies 1 + c = 3 \implies c = 2 \]

Thus, the \( n \)-th term is:
\[ T_n = \frac{3}{2}n^2 - \frac{1}{2}n + 2 \]

7. **Find the Sum of the First 15 Terms**:
The sum \( S_n \) of the first \( n \) terms is given by:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left( \frac{3}{2}k^2 - \frac{1}{2}k + 2 \right) \]

Using the formulas for the sums of \( k^2 \) and \( k \):
\[ S_n = \frac{3}{2} \cdot \frac{n(n+1)(2n+1)}{6} - \frac{1}{2} \cdot \frac{n(n+1)}{2} + 2n \]

For \( n = 15 \):
\[ S_{15} = \frac{3}{2} \cdot \frac{15 \cdot 16 \cdot 31}{6} - \frac{1}{2} \cdot \frac{15 \cdot 16}{2} + 30 \]

Calculate \( S_{15} \):
- \( \frac{15 \cdot 16 \cdot 31}{6} = 1240 \)
- \( S_{15} = \frac{3}{2} \cdot 1240 - \frac{1}{2} \cdot 120 + 30 \)
- \( S_{15} = 1860 - 60 + 30 = 1830 \)

8. **Set Equal to \( 15k \)**:
Given \( S_{15} = 15k \):
\[ 1830 = 15k \implies k = \frac{1830}{15} = 122 \]

### Final Answer:
Thus, the value of \( k \) is \( \boxed{122} \).