Answer
The proof is as shown below.
Work Step by Step
Here, $P(x)=\dfrac{x(x+1)}{2}$, so, $P(x+1)=\dfrac{(x+1)(x+2)}{2}$. Also, $P(1)=1$.
Now, we get
$\delta P=P(x+1)-P(x)$
$\,\,\,\,\,\;\,=\dfrac{(x+1)(x+2)}{2}-\dfrac{x(x+1)}{2}$
$\,\,\,\,\,\;\,=\dfrac{(x+1)}{2}\cdot\left(x+2-x\right)$
$\,\,\,\,\,\;\,=\dfrac{(x+1)}{2}\cdot\left(2\right)$
$\,\,\,\,\,\;\,=x+1$
Hence, proved.
Thus, $P(x+1)=P(x)+\delta P=P(x)+(x+1)$, so, we get
$P(2)=P(1)+2=1+2$;
$P(3)=P(2)+3=1+2+3$;
$P(4)=P(3)+4=1+2+3+4$;
Continuing in the same manner results, $P(n)=1+2+3+\cdots +n$.
Therefore, $1+2+3+\cdots+n=\dfrac{n(n+1)}{2}$.
