Question

Find the $4^{th}$ term of the series whose $n^{th}$ term is $\frac{2n+1}{n^2}$:

• A. $\frac{9}{16}$
• B. $\frac{7}{16}$
• C. $\frac{11}{16}$
• D. $\frac{16}{9}$

Answer 1

The correct option is A $\frac{9}{16}$

Given:

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

$\therefore t_4=\frac{2\times 4+1}{4^2}=\frac{9}{16}$

Hence, the answer is $\frac{9}{16}$.