2 + 4 + 6 + …+ 2n = n2 + n for all natural

According to the question,

P(n) is 2 + 4 + 6 + …+ 2n = n2 + n.

So, substituting different values for n, we get,

P(0) = 0 = 02 + 0 Which is true.

P(1) = 2 = 12 + 1 Which is true.

P(2) = 2 + 4 = 22 + 2 Which is true.

P(3) = 2 + 4 + 6 = 32 + 2 Which is true.

Let P(k) = 2 + 4 + 6 + …+ 2k = k2 + k be true;

So, we get,

⇒ P(k+1) is 2 + 4 + 6 + …+ 2k + 2(k+1) = k2 + k + 2k +2

= (k2 + 2k +1) + (k+1)

= (k + 1)2 + (k + 1)

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

2 + 4 + 6 + …+ 2n = n2 + n is true for all natural numbers n.