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Let $t_{n}$ be the sum of the first $n$ natural numbers, for $n > 0$. A number is called triangular if it is equal to $t_{n}$ for some $n$. Which of the following statements are true:

(i) There exists three successive triangular numbers whose product is a perfect square.

(ii) If the triangular number $t_{n}$ is a perfect square, then so is $t_{4n(n+1)}$.

(iii) The sum of the reciprocals of the first $n$ triangular numbers is less than $2$, i.e.

$\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+....+\frac{1}{t_{n}}<2$

1. $(i)$ only.
2. $(ii)$ only.
3. $(iii)$ only.
4. All of the above.
5. None of the above.
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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: $D$
by Veteran (56.8k points)
edited

(i) (x-1). x.(x+1)=p

i.e.x(x2-1) = p2

but (x2-1) itself is not a perfect square

So, There exists three successive triangular numbers whose product is a perfect square - FALSE

(ii)Similarly it is also false

(iii)1/1 +1/3+ 1/6+..........

=1+1/3(1+1/2+1/3.........)

<2, it is TRUE

Ans will be (C)

by Veteran (117k points)
0
if x is a triangular number, then x-1, x and x+1 are not the consecutive triangular numbers. they r simply consecutive integers.
0
ok