The formula for the sum of the first n natural numbers is given by the formula:

S = n/2 * (n + 1)

Here's how we derive it:

1. Let's denote the sum of the first n natural numbers as S. So, S = 1 + 2 + 3 + ... + n.
2. Now, let's write this sum backwards, i.e., S = n + (n-1) + (n-2) + ... + 1.
3. If we add these two equations, we get 2S = (n+1) + (n+1) + ... + (n+1) (n times). So, 2S = n*(n+1).
4. Therefore, S = n/2 * (n + 1).

This formula is a quadratic polynomial in n, as required.

Question

The formula for the sum of the first n natural numbers is given by the formula:

S = n/2 * (n + 1)

Here's how we derive it:

1. Let's denote the sum of the first n natural numbers as S. So, S = 1 + 2 + 3 + ... + n.
2. Now, let's write this sum backwards, i.e., S = n + (n-1) + (n-2) + ... + 1.
3. If we add these two equations, we get 2S = (n+1) + (n+1) + ... + (n+1) (n times). So, 2S = n*(n+1).
4. Therefore, S = n/2 * (n + 1).

This formula is a quadratic polynomial in n, as required.

Knowee AI · Accepted Answer

The formula for the sum of the first n natural numbers is given by the formula:

S = n/2 * (n + 1)

Here's how we derive it:

1. Let's denote the sum of the first n natural numbers as S. So, S = 1 + 2 + 3 + ... + n.
2. Now, let's write this sum backwards, i.e., S = n + (n-1) + (n-2) + ... + 1.
3. If we add these two equations, we get 2S = (n+1) + (n+1) + ... + (n+1) (n times). So, 2S = n*(n+1).
4. Therefore, S = n/2 * (n + 1).

This formula is a quadratic polynomial in n, as required.

Find a formula for the sum of the first n natural numbers. Assume that the formulais a quadratic polynomial in n.

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