Prove the following by using the principle of mathematical induction for all $n \in N$:
$1+2+3+\ldots+n < \frac{1}{8}(2n+1)^{2}$

Let the given statement be $P(n)$,i.e.,
$P(n): 1+2+3+\ldots+n < \frac{1}{8}(2n+1)^{2}$
Step $1$: Check for $n=1$:
$1 < \frac{1}{8}(2(1)+1)^{2} = \frac{9}{8} = 1.125$
Since $1 < 1.125$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some positive integer $k$,i.e.,
$1+2+\ldots+k < \frac{1}{8}(2k+1)^{2}$ ...........$(i)$
Step $3$: Prove $P(k+1)$ is true:
Consider the sum up to $(k+1)$:
$(1+2+\ldots+k) + (k+1) < \frac{1}{8}(2k+1)^{2} + (k+1)$
$= \frac{1}{8} \left\{ (2k+1)^{2} + 8(k+1) \right\}$
$= \frac{1}{8} \left\{ 4k^{2} + 4k + 1 + 8k + 8 \right\}$
$= \frac{1}{8} \left\{ 4k^{2} + 12k + 9 \right\}$
$= \frac{1}{8} (2k+3)^{2}$
$= \frac{1}{8} \{2(k+1)+1\}^{2}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,$P(n)$ is true for all $n \in N$.