Find the sum, `sum_(k=1) ^(20) (1+2+3+.....+k)`

Options

A. 1240
B. 1340
C. 1440
D. 1540

Explanation

To solve the problem of finding the sum \( \sum_{k=1}^{20} (1 + 2 + 3 + \ldots + k) \), we can break it down step by step.

### Step 1: Understand the Inner Sum
The expression \( 1 + 2 + 3 + \ldots + k \) represents the sum of the first \( k \) natural numbers. The formula for the sum of the first \( k \) natural numbers is given by:
\[ S_k = \frac{k(k + 1)}{2} \]

### Step 2: Rewrite the Summation
We can rewrite our original summation using the formula for \( S_k \):
\[ \sum_{k=1}^{20} (1 + 2 + 3 + \ldots + k) = \sum_{k=1}^{20} S_k = \sum_{k=1}^{20} \frac{k(k + 1)}{2} \]
This simplifies to:
\[ \sum_{k=1}^{20} S_k = \frac{1}{2} \sum_{k=1}^{20} k(k + 1) \]

### Step 3: Expand the Summation
We can expand \( k(k + 1) \) as follows:
\[ k(k + 1) = k^2 + k \]
Thus, we can split the summation:
\[ \sum_{k=1}^{20} k(k + 1) = \sum_{k=1}^{20} k^2 + \sum_{k=1}^{20} k \]

### Step 4: Use Known Formulas for Summations
We will use the formulas for the sum of the first \( n \) natural numbers and the sum of the squares of the first \( n \) natural numbers:
1. The sum of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
2. The sum of the squares of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

For \( n = 20 \):
- Calculate \( \sum_{k=1}^{20} k \):
\[ \sum_{k=1}^{20} k = \frac{20(20 + 1)}{2} = \frac{20 \times 21}{2} = 210 \]

- Calculate \( \sum_{k=1}^{20} k^2 \):
\[ \sum_{k=1}^{20} k^2 = \frac{20(20 + 1)(2 \times 20 + 1)}{6} = \frac{20 \times 21 \times 41}{6} \]
First, calculate \( 20 \times 21 = 420 \), then \( 420 \times 41 = 17220 \), and finally:
\[ \frac{17220}{6} = 2870 \]

### Step 5: Combine the Results
Now we can combine the results:
\[ \sum_{k=1}^{20} k(k + 1) = \sum_{k=1}^{20} k^2 + \sum_{k=1}^{20} k = 2870 + 210 = 3080 \]

### Step 6: Final Calculation
Now substitute back into our earlier expression:
\[ \sum_{k=1}^{20} (1 + 2 + 3 + \ldots + k) = \frac{1}{2} \sum_{k=1}^{20} k(k + 1) = \frac{1}{2} \times 3080 = 1540 \]

### Final Answer
Thus, the sum \( \sum_{k=1}^{20} (1 + 2 + 3 + \ldots + k) \) is:
\[ \boxed{1540} \]