Proof:

Let \[ P(n) \] be the statement:

\[ 2 + 4 + 6 + \dots + 2n = n(n+1) \]

Basic Step:

For \[ n=1 \]

\[ \text{LHS} = 2 \];
\[ \text{RHS} = n(n+1) \]
\[ = 1(1+1) \]
\[ = 2 \]

\[ \therefore P(1) \] is true

Inductive Step:

Assume \[ P(k) \] is true.

i.e., \[ 2 + 4 + 6 + \dots + 2k = k(k+1) \]

T.P \[ P(k+1) \text{ is true.} \]

i.e., T.P \[ 2 + 4 + 6 + \dots + 2(k+1) = (k+1)(k+2) \]

\[ \text{LHS}: \]

\[ 2 + 4 + 6 + \dots + 2(k+1) = 2 + 4 + 6 + \dots + 2k + 2(k+1) \]

\[= k(k+1) + 2(k+1)\]

\[= k^2 + k + 2k + 2 \]

\[= k^2 + 3k + 2 \]

\[= (k + 1)(k + 2) \]

\[= \text{RHS} \]

\[ \therefore P(k + 1) \] is true

\[ \therefore P(n) \] is true

\[ \therefore 2 + 4 + 6 + \dots + 2n = n (n+1) \forall n \]