2+4+6+….+2n=n(n+1) by mathematical induction
To prove the statement by mathematical induction, we need to show two things:
1. Base Case: Show that the statement holds true for n = 1.
2. Inductive Step: Assume that the statement is true for some positive integer k, and prove that it also holds for k+1.
Base Case (n=1):
When n = 1, the left-hand side of the equation is simply 2(1) = 2. The right-hand side is 1(1 + 1) = 2 as well. Therefore, the equation holds true for n = 1.
Inductive Step:
Assume that the equation holds true for some positive integer k. That is, 2 + 4 + 6 + ... + 2k = k(k + 1).
Now, let's consider the case for n = k + 1:
2 + 4 + 6 + ... + 2k + 2(k + 1)
Using the assumption, we can substitute k(k + 1) in place of the sum of the first k terms:
k(k + 1) + 2(k + 1)
Now, let's factor out (k + 1) from each term:
(k + 1)(k + 2)
This expression is equal to (k + 1)(k + 1 + 1), which is exactly k(k + 1). Therefore, the equation holds true for n = k + 1.
By mathematical induction, we have shown that the equation 2 + 4 + 6 + ... + 2n = n(n + 1) holds for all positive integers n.