Question

Prove that $1+2+3+.....+n=\frac{n(n+1)}{2}$. for $n$ being a natural numbers.

Answer 1

Let,

$p\left(n\right)=1+2+3+...\dots +n$

Now, $p\left(1\right)=1=\frac{1.(1+1)}{2}=1$

So, $p\left(1\right)$ is true

Let, us assume that $p\left(k\right)$ is true that is

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

We shall prove that $p(k+1)$ is true

Now,

$p(k+1)$

$=(1+2+.....+k)+(k+1)$

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

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

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

So, $p(k+1)$ is true.

Now by principle of mathematical induction curve $p\left(n\right)$ is true

$\forall n\in N$.