To Prove:

2 + 4 + 6 + 8 + …. + 2n = n(n + 1)

Steps to prove by mathematical induction:

Let P(n) be a statement involving the natural number n such that

(i) P(1) is true

(ii) P(k + 1) is true, whenever P(k) is true

Then P(n) is true for all n ϵ N

Therefore,

Let P(n): 2 + 4 + 6 + 8 + …. + 2n = n(n + 1)

Step 1:

P(1) = 1(1 + 1) = 1 × 2 = 2

Therefore, P(1) is true

Step 2:

Let P(k) is true Then,

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

Now,

2 + 4 + 6 + 8 + …. + 2k + 2(k + 1) = k(k + 1) + 2(k + 1)

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

= (k + 1) (k + 2)

= P(k + 1)

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

Hence, by the principle of mathematical induction, we have

2 + 4 + 6 + 8 + … + 2n = n(n + 1) for all n ϵ N