The sum of the first n even numbers is n(n+1).
Let's break down why this is the case. The first n even numbers can be represented as:
2, 4, 6, 8, ..., 2n
We can rewrite this sequence by factoring out a 2 from each term:
2(1), 2(2), 2(3), 2(4), ..., 2(n)
Now, we want to find the sum of this sequence:
S = 2(1) + 2(2) + 2(3) + ... + 2(n)
We can factor out the 2:
S = 2(1 + 2 + 3 + ... + n)
The sum of the first n natural numbers (1 + 2 + 3 + ... + n) is given by the formula:
Sum of first n natural numbers = n(n+1)/2
Substituting this back into our equation for the sum of the first n even numbers:
S = 2 * [n(n+1)/2]
The 2's cancel out, leaving us with:
S = n(n+1)
Therefore, the sum of the first n even numbers is indeed n(n+1).
Example:
Let's take n = 5. The first 5 even numbers are 2, 4, 6, 8, and 10.
Their sum is 2 + 4 + 6 + 8 + 10 = 30.
Using the formula, n(n+1) = 5(5+1) = 5(6) = 30. This confirms the formula.
