To prove this statement using mathematical induction, we need to show that it holds for the base case (when n = 1) and then show that if it holds for a general case (when n = k), it also holds for the next case (n = k+1).
Base Case (n = 1):
When n = 1, the sum 1+2+4+6...2n becomes 1. The right side of the equation becomes n(n+1) = 1(1+1) = 1(2) = 2. Since 1 is not equal to 2, the statement does not hold for n = 1.
However, if we assume that the statement holds for n = 2, we can proceed with the induction step. So let's assume that the sum 1+2+4+6...2n = n(n+1) holds for n = k:
Assumption (n = k):
1+2+4+6...2k = k(k+1)
Induction Step (n = k+1):
We need to prove that assuming the statement holds for n = k, it also holds for n = k+1. So let's consider the sum 1+2+4+6...2(k+1):
1+2+4+6...2(k+1) = 1+2+4+6...2k + 2(k+1)
Using our assumption (n = k), we can substitute the sum for n = k:
= k(k+1) + 2(k+1)
= (k+1)(k+2)
This shows that the statement holds for n = k+1. Hence, by mathematical induction, we have proven that for every natural number n, the sum 1+2+4+6...2n equals n(n+1).
1+2+4+6...2n=n(n+1) for every natural number proof by mathematical induction
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