Question

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

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is correct and Reason is incorrect

Answer 1

The correct option is D Assertion is correct and Reason is incorrect

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