The sum of the series
`3+8+16+27+41`…………… upto 20 terms is equal to

Options

A. 4230
B. 4430
C. 4330
D. 4500

Answer

To find the sum of the series \(3 + 8 + 16 + 27 + 41 + \ldots\) up to 20 terms, we will first determine the general term of the series and then calculate the sum.

### Step 1: Identify the pattern in the series
The given series is:
- \(T_1 = 3\)
- \(T_2 = 8\)
- \(T_3 = 16\)
- \(T_4 = 27\)
- \(T_5 = 41\)

Let's find the differences between consecutive terms:
- \(T_2 - T_1 = 8 - 3 = 5\)
- \(T_3 - T_2 = 16 - 8 = 8\)
- \(T_4 - T_3 = 27 - 16 = 11\)
- \(T_5 - T_4 = 41 - 27 = 14\)

The first differences are \(5, 8, 11, 14\). Now, let's find the second differences:
- \(8 - 5 = 3\)
- \(11 - 8 = 3\)
- \(14 - 11 = 3\)

Since the second differences are constant, the sequence is quadratic. We can express the \(n\)-th term as:
\[ T_n = an^2 + bn + c \]

### Step 2: Set up equations to find \(a\), \(b\), and \(c\)
Using the first three terms:
1. For \(n=1\): \(a(1)^2 + b(1) + c = 3 \quad \Rightarrow \quad a + b + c = 3\) (Equation 1)
2. For \(n=2\): \(a(2)^2 + b(2) + c = 8 \quad \Rightarrow \quad 4a + 2b + c = 8\) (Equation 2)
3. For \(n=3\): \(a(3)^2 + b(3) + c = 16 \quad \Rightarrow \quad 9a + 3b + c = 16\) (Equation 3)

### Step 3: Solve the system of equations
Subtract Equation 1 from Equation 2:
\[ (4a + 2b + c) - (a + b + c) = 8 - 3 \quad \Rightarrow \quad 3a + b = 5 \quad \text{(Equation 4)} \]

Subtract Equation 2 from Equation 3:
\[ (9a + 3b + c) - (4a + 2b + c) = 16 - 8 \quad \Rightarrow \quad 5a + b = 8 \quad \text{(Equation 5)} \]

Now, subtract Equation 4 from Equation 5:
\[ (5a + b) - (3a + b) = 8 - 5 \quad \Rightarrow \quad 2a = 3 \quad \Rightarrow \quad a = \frac{3}{2} \]

Substituting \(a\) back into Equation 4:
\[ 3\left(\frac{3}{2}\right) + b = 5 \quad \Rightarrow \quad \frac{9}{2} + b = 5 \quad \Rightarrow \quad b = 5 - \frac{9}{2} = \frac{10}{2} - \frac{9}{2} = \frac{1}{2} \]

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

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

### Step 4: Find the sum of the first 20 terms
The sum \(S_n\) of the first \(n\) terms is given by:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left(\frac{3}{2}k^2 + \frac{1}{2}k + 1\right) \]
\[ = \frac{3}{2} \sum_{k=1}^{n} k^2 + \frac{1}{2} \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]

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

Substituting these into the sum:
\[ S_n = \frac{3}{2} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{1}{2} \cdot \frac{n(n+1)}{2} + n \]
\[ = \frac{n(n+1)(2n+1)}{4} + \frac{n(n+1)}{4} + n \]
\[ = \frac{n(n+1)(2n+2)}{4} + n = \frac{n(n+1)(n+1)}{2} + n \]
\[ = \frac{n(n+1)^2}{2} + n \]

Now substituting \(n = 20\):
\[ S_{20} = \frac{20(21)(21)}{2} + 20 = 20 \cdot 21 \cdot 10.5 + 20 = 4410 + 20 = 4430 \]

### Final Answer
The sum of the series up to 20 terms is \(4430\).