Answer
$\frac{205}{24}$
Work Step by Step
$\sum_{j=1}^{4} \frac{6}{j^2}=\frac{6}{1^2}+\frac{6}{2^2}+\frac{6}{3^2}+\frac{6}{4^2}=\frac{6}{1}+\frac{6}{4}+\frac{6}{9}+\frac{6}{16}=\frac{144}{24}+\frac{36}{24}+\frac{16}{24}+\frac{9}{24}=\frac{205}{24}$
