Question

Evaluate: $\sum _{i=1}^{\infty }\frac{i^3}{3^i}$.

• A. $\frac{33}{8}$
• B. $\frac{35}{8}$
• C. $\frac{33}{7}$
• D. $\frac{35}{7}$

Answer 1

The correct option is A $\frac{33}{8}$

$S=\frac{1}{3}+\frac{8}{9}+\frac{27}{27}+\frac{64}{81}+\frac{128}{243}+....\infty \underset{–}{-\frac{1}{3}S=\frac{1}{9}+\frac{8}{27}+\frac{27}{81}+\frac{64}{243}+....\infty }\frac{2}{3}S=\frac{1}{3}+\frac{7}{9}+\frac{15}{27}+\frac{27}{81}+\frac{61}{243}+....\infty \underset{–}{-\frac{2}{9}S=\frac{1}{9}+\frac{7}{27}+\frac{19}{81}+\frac{37}{243}+....\infty }\frac{4}{9}S=\frac{1}{3}+\frac{6}{9}+\frac{12}{27}+\frac{18}{81}+\frac{24}{243}+....\infty \underset{–}{\frac{4}{27}S=\frac{1}{9}+\frac{6}{27}+\frac{12}{81}+\frac{18}{243}+....\infty }\frac{8}{27}S=\frac{1}{3}+\frac{8}{9}+\frac{6}{27}+\frac{6}{81}+\frac{6}{243}+....\infty$

$\frac{8}{27}S=\frac{8}{9}+\frac{6}{27}\frac{1}{(1-\frac{1}{3})}=\frac{8}{9}+\frac{6}{27}\times \frac{3}{2}=\frac{11}{9}S=\frac{33}{8}$