**Question:**

Using the principle of mathematical induction, prove each of the following for all n ϵ N:

$1+2+3+4+\ldots+n=1 / 2 n(n+1)$

**Solution:**

To Prove:

$1+2+3+4+\ldots+n=1 / 2 n(n+1)$

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 $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$

Therefore,

Let $P(n): 1+2+3+4+\ldots+n=1 / 2 n(n+1)$

Step 1:

P(1) = 1/2 1(1 + 1) = 1/2 × 2 = 1

Therefore, P(1) is true

Step 2

Let P(k) is true Then,

P(k): 1 + 2 + 3 + 4 + … + k = 1/2 k(k + 1)

Now,

1 + 2 + 3 + 4 + … + k + (k + 1) = 1/2 k(k + 1) + (k + 1)

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

= 1/2 (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

$1+2+3+4+\ldots+n=1 / 2 n(n+1)$ for all $n \in N$

Hence proved.