Question

Assertion (A) : If $S_{n}$ denotes the sum of $n$ terms of a series given by $S_{n} = \frac{n \left(n +1\right) \left(n +2\right)}{6} \forall n \ge1$ then $\lim_{n\to\infty}\sum\limits ^{n}_{r=1} \frac{1}{t_{r}} =4$
Reason (R) : $t_{n} = S_{n} - S_{n -1}$

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A) is true but (R) is false.
• D. (A) is false but (R) is true

Answer 1

Answer: (D)

$\because t_{n} =S_{n} -S_{n -1}$
$= \frac{n \left(n +1\right) \left(n +2\right)}{6}-\frac{\left(n -1\right) n\left(n +1\right)}{6}$
$t_{n} =\frac{1}{2} n\left(n +1\right)$
$\therefore \frac{1}{t_{n}} =\frac{2}{n\left(n +1\right)} =2 \left[\frac{1}{n} -\frac{1}{n +1}\right]$
$\therefore \sum\limits^{n}_{r =1} \frac{1}{t_{r}} =2$
$\left[\left(1-\frac{1}{2}\right) +\left(\frac{1}{2} -\frac{1}{3}\right) +....+\left(\frac{1}{n}-\frac{1}{n +1}\right)\right]$
$ =2 \left[1-\frac{1}{n +1}\right]$
$\therefore \lim_{n\to\infty} \sum\limits^{n}_{r =1} \frac{1}{t_{r}} =2$
$\lim_{n\to\infty} \left[1 -\frac{1}{n +1}\right] = 2\ne4$
$\therefore $ Assertion (A) is false but Reason (R) is true.