Question

If a sequence $\left\{a_n\right\}$ is defined as
$a_n=\{\begin{array}{c}\frac{1}{n},\text{when}n\text{is odd}\\ -\frac{1}{n^2},\text{when}n\text{is even}\end{array}$
where $n\in N$, then the sum of first four terms is

• A. $\frac{49}{48}$
• B. $\frac{47}{48}$
• C. $1$
• D. $2$

Answer 1

The correct option is A $\frac{49}{48}$

$a_n=\{\begin{array}{c}\frac{1}{n},\text{when}n\text{is odd}\\ -\frac{1}{n^2},\text{when}n\text{is even}\end{array}$
So,
$a_1=1,a_2=-\frac{1}{4},a_3=\frac{1}{3},a_4=-\frac{1}{16}$

Hence, the required sum
$=1-\frac{1}{4}+\frac{1}{3}-\frac{1}{16}=\frac{49}{48}$