Question

In a sequence, if $t_n=\frac{n^2-1}{n+1}$, then find the value of $S_6-S_3$.

• A. 3
• B. 6
• C. 12
• D. 10

Answer 1

Answer: (C)

12

Given, $t_n=\frac{n^2-1}{n+1}$

We know that $S_6$ is the sum of terms till $6th$ terms and $S_6$ is the sum of terms till $3rd$ term. SO, $S_6-S_3$ is the sum of $4th,5th,6th$ terms.

So, $t_4=\frac{4^2-1}{4+1}=\frac{15}{5}=3$

and $t_5=\frac{5^2-1}{5+1}=\frac{24}{6}=4$

and $t_6=\frac{6^2-1}{6+1}=\frac{35}{7}=5$

So, $S_6-S_3=t_4+t_5+t_6=3+4+5=12$