If `a_(1) = 2` and `a_(n) - a_(n-1) = 2n (n ge 2)`, find the value of `a_(1) + a_(2) + a_(3)+…+a_(20)`.

Options

A. 2960
B. 3080
C. 3560
D. 4120

Answer

To solve the problem, we need to find the value of \( a_1 + a_2 + a_3 + \ldots + a_{20} \) given that \( a_1 = 2 \) and \( a_n - a_{n-1} = 2n \) for \( n \geq 2 \).

### Step-by-Step Solution:

1. **Identify the first term:**
\[ a_1 = 2 \]

2. **Find \( a_2 \):**
Using the relation \( a_n - a_{n-1} = 2n \):
\[ a_2 - a_1 = 2 \times 2 \implies a_2 - 2 = 4 \implies a_2 = 6 \]

3. **Find \( a_3 \):**
\[ a_3 - a_2 = 2 \times 3 \implies a_3 - 6 = 6 \implies a_3 = 12 \]

4. **Find \( a_4 \):**
\[ a_4 - a_3 = 2 \times 4 \implies a_4 - 12 = 8 \implies a_4 = 20 \]

5. **Find \( a_5 \):**
\[ a_5 - a_4 = 2 \times 5 \implies a_5 - 20 = 10 \implies a_5 = 30 \]

6. **Observe the pattern:**
We can see that the differences \( a_n - a_{n-1} \) are \( 4, 6, 8, 10, \ldots \), which can be expressed as \( 2n \).

7. **General formula for \( a_n \):**
We can derive a general formula for \( a_n \):
\[ a_n = a_1 + \sum_{k=2}^{n} (a_k - a_{k-1}) = 2 + \sum_{k=2}^{n} 2k \]
The sum \( \sum_{k=2}^{n} 2k = 2 \sum_{k=2}^{n} k = 2 \left( \frac{n(n+1)}{2} - 1 \right) = n(n+1) - 2 \).

Hence, we have:
\[ a_n = 2 + n(n+1) - 2 = n(n+1) \]

8. **Calculate \( a_n \) for \( n = 1 \) to \( 20 \):**
\[ a_n = n(n + 1) \]
Therefore:
\[ a_1 = 1 \cdot 2 = 2 \]
\[ a_2 = 2 \cdot 3 = 6 \]
\[ a_3 = 3 \cdot 4 = 12 \]
\[ a_4 = 4 \cdot 5 = 20 \]
\[ a_5 = 5 \cdot 6 = 30 \]
\[ \vdots \]
\[ a_{20} = 20 \cdot 21 = 420 \]

9. **Sum \( a_1 + a_2 + \ldots + a_{20} \):**
\[ S = \sum_{n=1}^{20} a_n = \sum_{n=1}^{20} n(n + 1) \]
This can be split into:
\[ S = \sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} n \]

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

For \( N = 20 \):
\[ \sum_{n=1}^{20} n = \frac{20 \cdot 21}{2} = 210 \]
\[ \sum_{n=1}^{20} n^2 = \frac{20 \cdot 21 \cdot 41}{6} = 2870 \]

Therefore:
\[ S = 2870 + 210 = 3080 \]

### Final Answer:
\[ a_1 + a_2 + a_3 + \ldots + a_{20} = 3080 \]