by using methmatical induction , prove that the eqvation 2+4+6+....2n =n(n+1) is true for all positive integers

 To prove the equation 2 + 4 + 6 + ... + 2n = n(n + 1) for all positive integers using mathematical induction, we need to show that it holds for the base case (n = 1) and then demonstrate that if it holds for some arbitrary positive integer k, it also holds for k + 1.

**Step 1: Base Case**

Let's check the equation for the base case when n = 1:

2 = 1(1 + 1)

2 = 1(2)

2 = 2

The equation holds true for the base case.

**Step 2: Inductive Hypothesis**

Assume that the equation holds true for some arbitrary positive integer k, that is, assume that:

2 + 4 + 6 + ... + 2k = k(k + 1)

**Step 3: Inductive Step**

We need to show that if the equation holds for k, it also holds for k + 1.

We start with the left-hand side (LHS) of the equation:

2 + 4 + 6 + ... + 2k + 2(k + 1)

Using the assumption from the inductive hypothesis:

= k(k + 1) + 2(k + 1)

Factoring out (k + 1):

= (k + 1)(k + 2)

This matches the right-hand side (RHS) of the equation, which is n(n + 1) when n = k + 1:

= (k + 1)((k + 1) + 1)

= (k + 1)(k + 2)

Therefore, if the equation holds for k, it also holds for k + 1.

**Step 4: Conclusion**

By proving the base case and showing that if it holds for k, it also holds for k + 1, we have established that the equation 2 + 4 + 6 + ... + 2n = n(n + 1) is true for all positive integers by mathematical induction.