1+2+4+6...2n=n(n+1) for every natural number proof by mathematical induction
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).
To prove that the formula 1 + 2 + 4 + 6 + ... + 2n = n(n+1) holds true for every natural number n, we can use mathematical induction.
Step 1: Base Case
We start by proving that the formula is true for the smallest possible value of n, which is typically n = 1.
When n = 1:
The left-hand side of the formula is 1, and the right-hand side is 1(1+1) = 2. Since both sides are equal, the formula holds true for n = 1.
Step 2: Inductive Hypothesis
Assume that the formula holds true for some arbitrary value of n = k, where k is a positive integer.
Assume: 1 + 2 + 4 + 6 + ... + 2k = k(k+1)
Step 3: Inductive Step
We need to prove that the formula holds true for n = k+1, using the assumption from step 2.
Let us consider the left-hand side of the equation when n = k+1:
1 + 2 + 4 + 6 + ... + 2k + 2(k+1)
Recall that we assumed that 1 + 2 + 4 + 6 + ... + 2k = k(k+1). So, we can substitute this value in:
k(k+1) + 2(k+1)
Now, let's simplify:
= (k^2 + k) + (2k + 2)
= k^2 + 3k + 2
We can also represent (k+1)(k+2):
= (k+1)(k+2)
= k^2 + 3k + 2
Since both expressions are equal, we have proved that the formula holds true for n = k+1.
Step 4: Conclusion
Based on the principle of mathematical induction, we have established that if the formula is true for n = 1, and if it is true for some arbitrary value n = k, then it is also true for n = k+1.
Therefore, the formula 1 + 2 + 4 + 6 + ... + 2n = n(n+1) holds true for every natural number, as proved by mathematical induction.