Q.Using the principle of mathematical induction, prove that $1+2+3+\ldots\ldots+n = \dfrac{1}{2}n(n+1)$

Verifying the base case $n=1$ and the inductive step $P(k)\Rightarrow P(k+1)$ proves $1+2+\cdots+n=\dfrac{n(n+1)}{2}$ for all natural numbers $n$.

Let $P(n)$ be the statement:

$1+2+3+\cdots+n=\frac{n(n+1)}{2}$

Step 1 (Base case, $n=1$):

$\text{LHS}=1,\qquad \text{RHS}=\frac{1(1+1)}{2}=1$

LHS $=$ RHS, so $P(1)$ is true.

Step 2 (Inductive hypothesis): Assume $P(k)$ is true for some $k\in\mathbb{N}$:

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

Step 3 (Inductive step): We prove $P(k+1)$ is true, i.e. $1+2+\cdots+k+(k+1)=\dfrac{(k+1)(k+2)}{2}$.

Add $(k+1)$ to both sides of the hypothesis:

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