Question

Prove that: 1 + 2 + 3 + ......... + n = $\frac{n(n+1)}{2}$ i.e., the sum of the first n natural numbers is $\frac{n(n+1)}{2}$.

Answer 1

Let P(n) : 1 + 2 + 3 + ........ + n = $\frac{n(n+1)}{2}$

For n = 1,

LHS of P(n) = 1

RHS of P(n) =$\frac{1(1+1)}{2}1=1$

Since, LHS = RHS

$\Rightarrow$ P(n) is true for n = 1

Let P(n) be true for n = k, so

1 + 2 + 3 + ........ + k = $\frac{k(k+1)}{2}......\left(1\right)$

Now

(1 + 2 + 3 + ....... + k) + (k + 1)

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

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

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

$=\frac{(k+1)\left[\right(k+1)+1]}{2}$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all n $ϵ$ N

So, by the principle of mathematical induction

P(n) : 1 + 2 + 3 + ........ + n = $\frac{n(n+1)}{2}$ is true for all n $ϵ$ N.