The sum of 50 terms of the series `3+7+13+21+31+43+….` is equal to `S_(50)`, then the valueof `(S_(50))/(12500)` is

Answer

To solve the problem, we need to find the sum of the first 50 terms of the series \(3 + 7 + 13 + 21 + 31 + 43 + \ldots\) and then calculate the value of \(\frac{S_{50}}{12500}\).

### Step 1: Identify the pattern in the series
The given series is:
- First term \(a_1 = 3\)
- Second term \(a_2 = 7\)
- Third term \(a_3 = 13\)
- Fourth term \(a_4 = 21\)
- Fifth term \(a_5 = 31\)
- Sixth term \(a_6 = 43\)

To find a general term, let's look at the differences between consecutive terms:
- \(7 - 3 = 4\)
- \(13 - 7 = 6\)
- \(21 - 13 = 8\)
- \(31 - 21 = 10\)
- \(43 - 31 = 12\)

The differences are \(4, 6, 8, 10, 12\), which are increasing by \(2\). This indicates that the series is quadratic.

### Step 2: Find the general term \(a_n\)
Assuming the general term is of the form \(a_n = An^2 + Bn + C\), we can use the first few terms to set up equations.

1. For \(n=1\): \(A(1^2) + B(1) + C = 3\) → \(A + B + C = 3\) (Equation 1)
2. For \(n=2\): \(A(2^2) + B(2) + C = 7\) → \(4A + 2B + C = 7\) (Equation 2)
3. For \(n=3\): \(A(3^2) + B(3) + C = 13\) → \(9A + 3B + C = 13\) (Equation 3)

Now we have a system of equations:
- \(A + B + C = 3\) (1)
- \(4A + 2B + C = 7\) (2)
- \(9A + 3B + C = 13\) (3)

### Step 3: Solve the system of equations
Subtract Equation 1 from Equation 2:
\[ (4A + 2B + C) - (A + B + C) = 7 - 3 \]
\[ 3A + B = 4 \quad (Equation 4) \]

Subtract Equation 2 from Equation 3:
\[ (9A + 3B + C) - (4A + 2B + C) = 13 - 7 \]
\[ 5A + B = 6 \quad (Equation 5) \]

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

Substituting \(A = 1\) into Equation 4:
\[ 3(1) + B = 4 \quad \Rightarrow \quad B = 1 \]

Substituting \(A = 1\) and \(B = 1\) into Equation 1:
\[ 1 + 1 + C = 3 \quad \Rightarrow \quad C = 1 \]

Thus, the general term is:
\[ a_n = n^2 + n + 1 \]

### Step 4: Find \(S_{50}\)
The sum of the first \(n\) terms is:
\[ S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (k^2 + k + 1) \]
This can be separated into three sums:
\[ S_n = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]

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

For \(n = 50\):
\[ S_{50} = \frac{50(51)(101)}{6} + \frac{50(51)}{2} + 50 \]

Calculating each part:
1. \(\frac{50(51)(101)}{6} = 42925\)
2. \(\frac{50(51)}{2} = 1275\)
3. \(50 = 50\)

Adding these together:
\[ S_{50} = 42925 + 1275 + 50 = 44250 \]

### Step 5: Calculate \(\frac{S_{50}}{12500}\)
Now we calculate:
\[ \frac{S_{50}}{12500} = \frac{44250}{12500} = 3.54 \]

### Final Answer
Thus, the value of \(\frac{S_{50}}{12500}\) is \(3.54\).