Answer
$P(n)$ is true for all values of $n$.
Work Step by Step
Suppose that $P(n)=1+2+3+.......+n=\dfrac{n(n+1)}{2}$
1. Our aim is to find that $P(n)$ is true for $n=1$
$1=\dfrac{1(1+1)}{2} \implies 1=1$
2. Our aim is to find that $P(n)$ is true for $n=k$.This, it will also true for $n=k+1$
$1+2+3.....+k=\dfrac{k(k+1)}{2} \\ \dfrac{k(k+1)}{2} +(k+1)=\dfrac{k^2+3k+2}{2}$
This yields: $ \dfrac{k^2+3k+2}{2}=\dfrac{k^2+3k+2}{2}$
Therefore, $P(n)$ is true for all values of $n$.
