Find the sum of n terms of the series `3+7+14+24+37+………`

Answer

To find the sum of the first \( n \) terms of the series \( 3 + 7 + 14 + 24 + 37 + \ldots \), we will first identify the \( n \)-th term of the series and then derive the sum.

### Step 1: Identify the pattern in the series
Let's denote the \( n \)-th term of the series as \( t_n \). The given terms are:
- \( t_1 = 3 \)
- \( t_2 = 7 \)
- \( t_3 = 14 \)
- \( t_4 = 24 \)
- \( t_5 = 37 \)

To find a pattern, we can calculate the differences between successive terms:
- \( t_2 - t_1 = 7 - 3 = 4 \)
- \( t_3 - t_2 = 14 - 7 = 7 \)
- \( t_4 - t_3 = 24 - 14 = 10 \)
- \( t_5 - t_4 = 37 - 24 = 13 \)

The differences are \( 4, 7, 10, 13 \), which form an arithmetic progression (AP) with a common difference of \( 3 \).

### Step 2: Find the general term \( t_n \)
Since the first differences form an AP, we can express the \( n \)-th term as:
\[ t_n = t_1 + \text{(sum of first (n-1) terms of the AP)} \]
The first term of the AP is \( 4 \) and the common difference is \( 3 \).

The sum of the first \( n-1 \) terms of an AP is given by:
\[ S_{n-1} = \frac{(n-1)}{2} \times [2a + (n-2)d] \]
where \( a = 4 \) and \( d = 3 \).

Substituting the values:
\[ S_{n-1} = \frac{(n-1)}{2} \times [2 \times 4 + (n-2) \times 3] = \frac{(n-1)}{2} \times [8 + 3n - 6] = \frac{(n-1)}{2} \times (3n + 2) \]

Thus, we have:
\[ t_n = 3 + S_{n-1} = 3 + \frac{(n-1)(3n + 2)}{2} \]

### Step 3: Simplify \( t_n \)
Now, we simplify \( t_n \):
\[ t_n = 3 + \frac{(n-1)(3n + 2)}{2} \]
\[ = 3 + \frac{3n^2 - 3n + 2n - 2}{2} \]
\[ = 3 + \frac{3n^2 - n - 2}{2} \]
\[ = \frac{6 + 3n^2 - n - 2}{2} \]
\[ = \frac{3n^2 - n + 4}{2} \]

### Step 4: Find the sum of the first \( n \) terms \( S_n \)
The sum of the first \( n \) terms \( S_n \) can be expressed as:
\[ S_n = \sum_{k=1}^{n} t_k = \sum_{k=1}^{n} \left( \frac{3k^2 - k + 4}{2} \right) \]
\[ = \frac{1}{2} \left( 3 \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k + 4n \right) \]

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

Substituting these into the equation:
\[ S_n = \frac{1}{2} \left( 3 \cdot \frac{n(n + 1)(2n + 1)}{6} - \frac{n(n + 1)}{2} + 4n \right) \]
\[ = \frac{1}{2} \left( \frac{n(n + 1)(2n + 1)}{2} - \frac{n(n + 1)}{2} + 4n \right) \]
\[ = \frac{1}{2} \left( \frac{n(n + 1)(2n + 1 - 1) + 8n}{2} \right) \]
\[ = \frac{1}{4} \left( n(n + 1)(2n) + 8n \right) \]
\[ = \frac{n(n + 1)(2n + 8)}{4} \]

### Final Result
Thus, the sum of the first \( n \) terms of the series is:
\[ S_n = \frac{n(n + 1)(2n + 8)}{4} \]