**2 + 4 + 6+… + 2n = n2 + n, for all natural numbers .**

###### Answers:

Let P(n) :2 + 4 + 6+ …+2 n = n^{2} + n

P(l): 2 = l^{2} + 1 = 2, which is true

Hence, P(l) is true.

Let us assume that P(n) is true for some natural number n = k.

∴ P(k): 2 + 4 + 6 + .,.+2k = k^{2} + k (i)

Now, we have to prove that P(k + 1) is true.

P(k + l):2 + 4 + 6 + 8+ …+2k+ 2 (k +1)

= k^{2} + k + 2(k+ 1) [Using (i)]

= k^{2} + k + 2k + 2

= k^{2} + 2k+1+k+1

= (k + 1)^{2} + k+ 1

Hence, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for any natural number n.