For all `n in N, Sigma n`

Options

A. `lt((2n+1)^(2))/(8)`
B. `gt((2n+1)^(2))/(8)`
C. `=((2n+1)^(2))/(8)`
D. none of these

Answer

To find the sum of the first \( n \) natural numbers, we can use the following step-by-step approach:

### Step 1: Define the Sum
Let \( S_n \) be the sum of the first \( n \) natural numbers:
\[ S_n = 1 + 2 + 3 + \ldots + n \]

### Step 2: Write the Sum in Summation Notation
We can express this sum using summation notation:
\[ S_n = \sum_{k=1}^{n} k \]

### Step 3: Reverse the Order of the Sum
Now, we can write the same sum in reverse order:
\[ S_n = n + (n-1) + (n-2) + \ldots + 1 \]

### Step 4: Add the Two Equations
Next, we add the two expressions for \( S_n \):
\[ S_n + S_n = (1 + 2 + 3 + \ldots + n) + (n + (n-1) + (n-2) + \ldots + 1) \]
This simplifies to:
\[ 2S_n = (1+n) + (2+(n-1)) + (3+(n-2)) + \ldots + (n+1) \]

### Step 5: Group the Terms
Notice that each pair sums to \( n + 1 \):
\[ 2S_n = (n + 1) + (n + 1) + (n + 1) + \ldots + (n + 1) \]
There are \( n \) such pairs.

### Step 6: Simplify the Equation
Thus, we can write:
\[ 2S_n = n(n + 1) \]

### Step 7: Solve for \( S_n \)
Now, divide both sides by 2 to solve for \( S_n \):
\[ S_n = \frac{n(n + 1)}{2} \]

### Conclusion
The sum of the first \( n \) natural numbers is given by:
\[ S_n = \frac{n(n + 1)}{2} \]