Sum up the following series to n terms:`3+7+14+24+37+…………..`

Answer

To find the sum of the series \(3 + 7 + 14 + 24 + 37 + \ldots\) up to \(n\) terms, we can follow these steps:

### Step 1: Identify the nth term
Let's denote the nth term of the series as \(T_n\). The given series is:
- \(T_1 = 3\)
- \(T_2 = 7\)
- \(T_3 = 14\)
- \(T_4 = 24\)
- \(T_5 = 37\)

### Step 2: Find the differences between consecutive terms
Now, let's find the differences between consecutive 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 \(4, 7, 10, 13\).

### Step 3: Find the second differences
Next, we find the differences of these differences:
- \(7 - 4 = 3\)
- \(10 - 7 = 3\)
- \(13 - 10 = 3\)

Since the second differences are constant, this indicates that the nth term \(T_n\) is a quadratic function of \(n\).

### Step 4: Assume a quadratic form for \(T_n\)
We can assume that:
\[ T_n = an^2 + bn + c \]
We need to find the coefficients \(a\), \(b\), and \(c\).

### Step 5: Set up equations using known terms
Using the first three terms:
1. For \(n = 1\): \(a(1^2) + b(1) + c = 3 \Rightarrow a + b + c = 3\)
2. For \(n = 2\): \(a(2^2) + b(2) + c = 7 \Rightarrow 4a + 2b + c = 7\)
3. For \(n = 3\): \(a(3^2) + b(3) + c = 14 \Rightarrow 9a + 3b + c = 14\)

### Step 6: Solve the system of equations
We have the following system of equations:
1. \(a + b + c = 3\) (1)
2. \(4a + 2b + c = 7\) (2)
3. \(9a + 3b + c = 14\) (3)

Subtract (1) from (2):
\[ (4a + 2b + c) - (a + b + c) = 7 - 3 \Rightarrow 3a + b = 4 \quad (4) \]

Subtract (2) from (3):
\[ (9a + 3b + c) - (4a + 2b + c) = 14 - 7 \Rightarrow 5a + b = 7 \quad (5) \]

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

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

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

### Step 7: Write the nth term
Thus, the nth term is:
\[ T_n = \frac{3}{2}n^2 - \frac{1}{2}n + 1 \]

### Step 8: Find the sum \(S_n\)
To find the sum \(S_n\) of the first \(n\) terms:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left(\frac{3}{2}k^2 - \frac{1}{2}k + 1\right) \]

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:
\[ S_n = \frac{3}{2} \cdot \frac{n(n+1)(2n+1)}{6} - \frac{1}{2} \cdot \frac{n(n+1)}{2} + n \]

### Step 9: Simplify \(S_n\)
Simplifying this expression:
\[ S_n = \frac{3n(n+1)(2n+1)}{12} - \frac{n(n+1)}{4} + n \]
\[ = \frac{3n(n+1)(2n+1) - 3n(n+1) + 12n}{12} \]
\[ = \frac{3n(n+1)(2n+1 - 1) + 12n}{12} \]
\[ = \frac{3n(n+1)(2n) + 12n}{12} \]
\[ = \frac{3n(n+1)(2n) + 12n}{12} \]

### Final Result
Thus, the sum of the series up to \(n\) terms is:
\[ S_n = \frac{3n^3 + 3n^2 + 12n}{12} \]