Question

Assertion (A) : If $S_n$ denotes the sum of $n$ terms of a series given by $S_n=\frac{n\left(n+1\right)\left(n+2\right)}{6}\forall n\ge 1$ then ${\lim }_{n\to \infty }\sum _{r=1}^n\frac{1}{t_r}=4$
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)

$\because t_n=S_n-S_{n-1}$
$=\frac{n\left(n+1\right)\left(n+2\right)}{6}-\frac{\left(n-1\right)n\left(n+1\right)}{6}$
$t_n=\frac{1}{2}n\left(n+1\right)$
$\therefore \frac{1}{t_n}=\frac{2}{n\left(n+1\right)}=2\left[\frac{1}{n}-\frac{1}{n+1}\right]$
$\therefore \sum _{r=1}^n\frac{1}{t_r}=2$
$\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+....+\left(\frac{1}{n}-\frac{1}{n+1}\right)\right]$
$=2\left[1-\frac{1}{n+1}\right]$
$\therefore {\lim }_{n\to \infty }\sum _{r=1}^n\frac{1}{t_r}=2$
${\lim }_{n\to \infty }\left[1-\frac{1}{n+1}\right]=2\ne 4$
$\therefore$ Assertion (A) is false but Reason (R) is true.