Question

$\frac{7}{2^3}+\frac{19}{6^3}+\frac{37}{{12}^3}+\frac{61}{{20}^3}+......\infty$.

Answer 1

$\frac{7}{2^3}+\frac{19}{6^3}+\frac{37}{{12}^3}+\frac{61}{{20}^3}$

$\sum _{r=1}^{\infty }\frac{3r^2+3r+1}{\left(r^2+r\right)3}=\frac{3{\left(r+1\right)}^3-r^3}{r^3{\left(r+1\right)}^3}$

$=\sum _{r=1}^{\infty }\frac{1}{r^3}-\frac{1}{{\left(r+1\right)}^3}$

$=\left(\frac{1}{1^3}-\frac{1}{2^3}\right)+\left(\frac{1}{2^3}-\frac{1}{3^3}\right)$

$=1.$