The sum of the first 20 terms of the series 5 + 11 + 19 + 29 + 41 + … is

Options

A. 3520
B. 3450
C. 3250
D. 3420

Answer

To find the sum of the first 20 terms of the series 5 + 11 + 19 + 29 + 41 + ..., we first need to determine the general term of the series.

### Step 1: Identify the pattern in the series
The given series is:
- 1st term (T1) = 5
- 2nd term (T2) = 11
- 3rd term (T3) = 19
- 4th term (T4) = 29
- 5th term (T5) = 41

Let's look at the differences between consecutive terms:
- T2 - T1 = 11 - 5 = 6
- T3 - T2 = 19 - 11 = 8
- T4 - T3 = 29 - 19 = 10
- T5 - T4 = 41 - 29 = 12

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

### Step 2: Assume a quadratic form for the nth term
We assume the nth term can be expressed as:
\[ T_n = an^2 + bn + c \]

### Step 3: Set up equations using known terms
Using the first three terms, we can set up the following equations:
1. For n = 1: \( a(1^2) + b(1) + c = 5 \) → \( a + b + c = 5 \) (Equation 1)
2. For n = 2: \( a(2^2) + b(2) + c = 11 \) → \( 4a + 2b + c = 11 \) (Equation 2)
3. For n = 3: \( a(3^2) + b(3) + c = 19 \) → \( 9a + 3b + c = 19 \) (Equation 3)

### Step 4: Solve the system of equations
Now we can solve these equations step by step.

**Subtract Equation 1 from Equation 2:**
\[ (4a + 2b + c) - (a + b + c) = 11 - 5 \]
This simplifies to:
\[ 3a + b = 6 \quad \text{(Equation 4)} \]

**Subtract Equation 2 from Equation 3:**
\[ (9a + 3b + c) - (4a + 2b + c) = 19 - 11 \]
This simplifies to:
\[ 5a + b = 8 \quad \text{(Equation 5)} \]

**Now, subtract Equation 4 from Equation 5:**
\[ (5a + b) - (3a + b) = 8 - 6 \]
This gives us:
\[ 2a = 2 \quad \Rightarrow \quad a = 1 \]

### Step 5: Find b and c
Substituting \( a = 1 \) back into Equation 4:
\[ 3(1) + b = 6 \quad \Rightarrow \quad b = 3 \]

Now substitute \( a \) and \( b \) into Equation 1:
\[ 1 + 3 + c = 5 \quad \Rightarrow \quad c = 1 \]

### Step 6: Write the general term
Now we have:
\[ T_n = n^2 + 3n + 1 \]

### Step 7: Find the sum of the first 20 terms
The sum of the first n terms \( S_n \) can be calculated using:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + 3k + 1) \]
This can be split into three separate sums:
\[ S_n = \sum_{k=1}^{n} k^2 + 3\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]

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

For \( n = 20 \):
1. \( \sum_{k=1}^{20} k^2 = \frac{20(21)(41)}{6} = 2870 \)
2. \( 3\sum_{k=1}^{20} k = 3 \cdot \frac{20(21)}{2} = 630 \)
3. \( \sum_{k=1}^{20} 1 = 20 \)

### Step 8: Combine the results
Now combine these results:
\[ S_{20} = 2870 + 630 + 20 = 3520 \]

### Final Answer
The sum of the first 20 terms of the series is **3520**.