Find the sum `3+7+14+24+37+…..20 ` terms

Explanation

To find the sum of the series \(3 + 7 + 14 + 24 + 37 + \ldots\) up to 20 terms, we first need to identify the pattern in the series and then derive a formula for the \(n\)-th term.

### Step 1: Identify the pattern in the series
The given series is:
- \(T_1 = 3\)
- \(T_2 = 7\)
- \(T_3 = 14\)
- \(T_4 = 24\)
- \(T_5 = 37\)

Let's calculate 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 first differences are \(4, 7, 10, 13\). Now, let's calculate the second differences:
- \(7 - 4 = 3\)
- \(10 - 7 = 3\)
- \(13 - 10 = 3\)

Since the second differences are constant, this indicates that the \(n\)-th term can be expressed as a quadratic function of \(n\):
\[ T_n = an^2 + bn + c \]

### Step 2: Set up equations to find \(a\), \(b\), and \(c\)
Using the first three terms of the series, we can set up the following equations:
1. For \(n=1\): \(T_1 = a(1^2) + b(1) + c = 3 \Rightarrow a + b + c = 3\)
2. For \(n=2\): \(T_2 = a(2^2) + b(2) + c = 7 \Rightarrow 4a + 2b + c = 7\)
3. For \(n=3\): \(T_3 = a(3^2) + b(3) + c = 14 \Rightarrow 9a + 3b + c = 14\)

Now we have the system of equations:
1. \(a + b + c = 3\) (Equation 1)
2. \(4a + 2b + c = 7\) (Equation 2)
3. \(9a + 3b + c = 14\) (Equation 3)

### Step 3: Solve the system of equations
Subtract Equation 1 from Equation 2:
\[ (4a + 2b + c) - (a + b + c) = 7 - 3 \Rightarrow 3a + b = 4 \quad \text{(Equation 4)} \]

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

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

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

Now substitute \(a\) and \(b\) back into Equation 1 to find \(c\):
\[ \frac{3}{2} - \frac{1}{2} + c = 3 \Rightarrow 2 + c = 3 \Rightarrow c = 1 \]

### Step 4: Write the formula for \(T_n\)
Now we have:
\[ T_n = \frac{3}{2}n^2 - \frac{1}{2}n + 1 \]

### Step 5: Calculate the sum of the first 20 terms
The sum \(S_n\) of the first \(n\) terms can be expressed as:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left(\frac{3}{2}k^2 - \frac{1}{2}k + 1\right) \]
This can be separated into three sums:
\[ S_n = \frac{3}{2} \sum_{k=1}^{n} k^2 - \frac{1}{2} \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]

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

Substituting \(n = 20\):
\[ S_{20} = \frac{3}{2} \cdot \frac{20(21)(41)}{6} - \frac{1}{2} \cdot \frac{20(21)}{2} + 20 \]

Calculating each term:
1. \(\frac{3}{2} \cdot \frac{20 \cdot 21 \cdot 41}{6} = \frac{3 \cdot 20 \cdot 21 \cdot 41}{12} = \frac{3 \cdot 20 \cdot 21 \cdot 41}{12}\)
2. \(\frac{1}{2} \cdot \frac{20 \cdot 21}{2} = \frac{20 \cdot 21}{4} = 105\)
3. The last term is simply \(20\).

Now, calculating these values:
\[ S_{20} = \frac{3 \cdot 20 \cdot 21 \cdot 41}{12} - 105 + 20 \]

Calculating \(3 \cdot 20 \cdot 21 \cdot 41\):
\[ = 3 \cdot 20 = 60 \\ 60 \cdot 21 = 1260 \\ 1260 \cdot 41 = 51660 \]
Now divide by 12:
\[ \frac{51660}{12} = 4305 \]
So,
\[ S_{20} = 4305 - 105 + 20 = 4200 \]

### Final Answer
Thus, the sum of the first 20 terms is:
\[ \boxed{4200} \]