A New Approach to Finding the Sum of the First n Natural Numbers
Posted on: March 9, 2025
We all know that the sum of the first n natural numbers can be found using the formula:
\[ S_n = \frac{n(n+1)}{2} \]
Here, we are going to explore a different approach to finding the sum of the firstn natural numbers, but this method applies only to even numbers.
We can rewrite the sum 1 + 2 + 3 + 4 + ... in the following pattern:
\[ -1^2 + 2^2 - 3^2 + 4^2 - 5^2 + 6^2 \dots \]
How does this work?
- \( 1+2 = 3 = -1^2 + 2^2 \)
- \( 3+4 = 7 = -3^2 + 4^2 \)
In general, this follows the known formula:
\[ (n+1)^2 - n^2 = 2n+1 \]
From this, we can observe a pattern:
\[ (2(1)+1) + (2(3)+1) + (2(5)+1) = 1 + 2 + 3 + 4 + 5 + 6 = 21 = S_6 \]
To generalize this sequence:
\[ \sum_{i=0}^{\frac{n}{2}-1} \left( 4i+3 \right) \]
Let's use this formula to compute \( S_6 \):
- n = 6, so we iterate from i = 0 to 2
- \( S_6 = (4(0) + 3) + (4(1) + 3) + (4(2) + 3) \)
- \( S_6 = 3 + 7 + 11 = 21 \)