Find the sum of the series `1+ (1+2) + (1+2+3) + (1+2 + 3 +4)+ … + (1+2+3+… + 20)`.

Options

A. 1470
B. 1540
C. 1610
D. 1370

Explanation

To find the sum of the series \( S = 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + \ldots + (1 + 2 + 3 + \ldots + 20) \), we can break it down step by step.

### Step 1: Identify the terms in the series
The series consists of sums of the first \( n \) natural numbers for \( n \) ranging from 1 to 20. We can denote the \( n \)-th term of the series as \( T_n \), where:
\[ T_n = 1 + 2 + 3 + \ldots + n \]

### Step 2: Use the formula for the sum of the first \( n \) natural numbers
The sum of the first \( n \) natural numbers is given by the formula:
\[ T_n = \frac{n(n + 1)}{2} \]

### Step 3: Write the entire series in terms of \( T_n \)
Thus, the sum \( S \) can be expressed as:
\[ S = T_1 + T_2 + T_3 + \ldots + T_{20} \]
Substituting the formula for \( T_n \):
\[ S = \frac{1(1 + 1)}{2} + \frac{2(2 + 1)}{2} + \frac{3(3 + 1)}{2} + \ldots + \frac{20(20 + 1)}{2} \]

### Step 4: Factor out the common term
We can factor out \( \frac{1}{2} \) from the sum:
\[ S = \frac{1}{2} \left( 1(1 + 1) + 2(2 + 1) + 3(3 + 1) + \ldots + 20(20 + 1) \right) \]

### Step 5: Simplify the expression inside the parentheses
Now we need to compute:
\[ \sum_{n=1}^{20} n(n + 1) = \sum_{n=1}^{20} (n^2 + n) \]
This can be separated into two sums:
\[ \sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} n \]

### Step 6: Use the formulas for the sums of squares and natural numbers
The formulas for these sums are:
- The sum of the first \( n \) natural numbers:
\[ \sum_{n=1}^{N} n = \frac{N(N + 1)}{2} \]
- The sum of the squares of the first \( n \) natural numbers:
\[ \sum_{n=1}^{N} n^2 = \frac{N(N + 1)(2N + 1)}{6} \]

### Step 7: Calculate the sums for \( N = 20 \)
For \( N = 20 \):
\[ \sum_{n=1}^{20} n = \frac{20(21)}{2} = 210 \]
\[ \sum_{n=1}^{20} n^2 = \frac{20(21)(41)}{6} = 2870 \]

### Step 8: Combine the results
Now substituting back into our expression:
\[ \sum_{n=1}^{20} n(n + 1) = 2870 + 210 = 3080 \]

### Step 9: Multiply by \( \frac{1}{2} \)
Finally, we compute \( S \):
\[ S = \frac{1}{2} \times 3080 = 1540 \]

### Final Answer
Thus, the sum of the series is:
\[ \boxed{1540} \]