To prove the statement:

$2 + 4 + 6 + \cdots + 2n = n(1 + n)$

We can proceed by mathematical induction.

Step 1: Base Case (n = 1)

For $n = 1$:

The left-hand side (LHS) is: $2 = 2$

The right-hand side (RHS) is: $n(1 + n) = 1(1 + 1) = 2$

Since LHS = RHS, the base case holds.

Step 2: Inductive Hypothesis

Assume that the statement holds for some $n = k$, i.e., $2 + 4 + 6 + \cdots + 2k = k(1 + k)$ This is the inductive hypothesis.

Step 3: Inductive Step

We need to show that the statement also holds for $n = k + 1$. That is, we want to prove: $2 + 4 + 6 + \cdots + 2k + 2(k + 1) = (k + 1)(1 + (k + 1))$

From the inductive hypothesis, we know: $2 + 4 + 6 + \cdots + 2k = k(1 + k)$

Now, add $2(k + 1)$ to both sides: $k(1 + k) + 2(k + 1)$

Simplify the right-hand side: