Using the principle of mathematical induction, prove each of the following
Question:

Using the principle of mathematical induction, prove each of the following for all $n \in \bar{N}$ :

2 + 4 + 6 + 8 + …. + 2n = n(n + 1)

Solution:

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 $n \in N$

Therefore,

Let $P(n): 2+4+6+8+\ldots .+2 n=n(n+1)$

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,

$2+4+6+8+\ldots+2 k+2(k+1)=k(k+1)+2(k+1)$

$=\mathrm{k}(\mathrm{k}+1)+2(\mathrm{k}+1)$

$=(\mathrm{k}+1)(\mathrm{k}+2)$

$=\mathrm{P}(\mathrm{k}+1)$

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

Hence, by the principle of mathematical induction, we have

$2+4+6+8+\ldots+2 n=n(n+1)$ for all $n \in N$