Triangular number, $t_n = \dfrac{n(n+1)}{2}$
Product of three consecutive Triangular numbers , $t_m \times t_{m+1} \times t_{m+2}$
$= \dfrac{m(m+1)}{2} \times \dfrac{(m+1)(m+2)}{2} \times \dfrac{(m+2)(m+3)}{2}$
$= \left(\dfrac{m(m+1)}{2}\right)^2 \times \left(\dfrac{m(m+3)}{2}\right)$
At $m= 3,$ $t_3 \times t_4 \times t_5$ is a perfect square.
(i) is True.
$t_{4n(n+1)} = t_{4n^2+4n} = \dfrac{(4n^2 +4n)(4n^2+4n+1)}{2}$
$\quad \quad =4 \times (2n+1)^2 \times \dfrac{n(n+1)}{2} = 2^2 \times (2n+1)^2 \times t_n$
If $t_n$ is a perfect square, then $t_{4n(n+1)}$ is also a perfect square
(ii) is True
$\dfrac{1}{1} + \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{10}+\ldots +\dfrac{1}{t_n}$
$\quad =\dfrac{2}{1.(1+1)} + \dfrac{2}{2.(2+1)} + \dfrac{2}{3.(3+1)} + \dfrac{2}{4.(4+1)}+\ldots +\dfrac{2}{n.(n+1)}$
$\quad = 2 \times \left(\dfrac{1}{1.(1+1)} + \dfrac{1}{2.(2+1)} + \dfrac{1}{3.(3+1)} + \dfrac{1}{4.(4+1)}+\ldots +\dfrac{1}{n.(n+1)}\right)$
$\quad = 2 \times \left(\dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} + \dfrac{1}{4.5}+\ldots +\dfrac{1}{n.(n+1)}\right)$
$\quad =2 \times \left(\left(\dfrac{1}{1} -\dfrac{1}{2}\right)+ \left(\dfrac{1}{2} -\dfrac{1}{3}\right)+ \left(\dfrac{1}{3} -\dfrac{1}{4}\right)+ \left(\dfrac{1}{4} -\dfrac{1}{5}\right)+\ldots +\left(\dfrac{1}{n}-\dfrac{1}{(n+1)}\right)\right)$
$\quad =2 \times \left(1 -\dfrac{1}{(n+1)}\right)$
$\quad =2 \times \dfrac{n}{n+1}$
for any $n>0$, $\dfrac{n}{n+1}$ will be $< 1$, so $2 \times \dfrac{n}{n+1}$ will be $<2$.
So, $\dfrac{1}{1} + \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{10}+\ldots+\dfrac{1}{t_n} < 2$
(iii) is also True.
Correct Answer: $D$