Question

Prove that $1+2+3+.....+n=\left[\frac{n(n+1)}{2}\right]$ for $n$ being a natural number.

Answer 1

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

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

$p\left(1\right)$ is true.

Let us assume $p\left(k\right)$ is true, that is $1+2+3+.........+k=\frac{k\left(k+1\right)}{2}$……………………..$\left(i\right)$

Now, we shall prove that $p\left(k+1\right)$ is true.

$p\left(k+1\right)$$=\left(1+2+...........+k\right)+\left(k+1\right)$

$=\frac{k\left(k+1\right)}{2}+\left(k+1\right)$………………………………(from $\left(i\right)$)

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

$=\frac{k^2+k+2k+2}{2}$

$=\frac{k^2+3k+2}{2}$

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

$=\frac{\left(k+1\right)\left(k+1+1\right)}{2}$………………….(It is of the form $\frac{n\left(n+1\right)}{2}$)

Therefore, $p\left(k+1\right)$ is true.

Now by the principle of mathematical induction, $p\left(n\right)$ is true for all $n\in N$.

Hence, $1+2+3+.....+n=\left[\frac{n(n+1)}{2}\right]$ for $n$ being a natural number.