The given series is

\(\sum_{n=0}^\infty((\frac{5}{2^n})-(\frac{1}{3^n}))\)

Rewrite the given series as shown below:

\(\sum_{n=0}^\infty((\frac{5}{2^n})-(\frac{1}{3^n}))=\sum_{n=0}^\infty(\frac{5}{2^n})-\sum_{n=0}^\infty(\frac{1}{3^n})\)

From above it can be observed that given series is the sum of two geometric series. Hence, the sum of series is obtained as,

\(\sum_{n=0}^\infty((\frac{5}{2^n})-(\frac{1}{3^n}))=\sum_{n=0}^\infty(\frac{5}{2^n})-\sum_{n=0}^\infty(\frac{1}{3^n})\)

\(=5\sum_{n=0}^\infty(\frac{1}{2^n})-\sum_{n=0}^\infty(\frac{1}{3^n})\)

\(=5(\frac{\frac12}{1-\frac12})-(\frac{\frac13}{1-\frac13})\)

\(=5(\frac{\frac12}{\frac12})-(\frac{\frac13}{\frac23})\)

\(=5(1)-(\frac12)\)

\(=\frac92\)

\(\sum_{n=0}^\infty((\frac{5}{2^n})-(\frac{1}{3^n}))\)

Rewrite the given series as shown below:

\(\sum_{n=0}^\infty((\frac{5}{2^n})-(\frac{1}{3^n}))=\sum_{n=0}^\infty(\frac{5}{2^n})-\sum_{n=0}^\infty(\frac{1}{3^n})\)

From above it can be observed that given series is the sum of two geometric series. Hence, the sum of series is obtained as,

\(\sum_{n=0}^\infty((\frac{5}{2^n})-(\frac{1}{3^n}))=\sum_{n=0}^\infty(\frac{5}{2^n})-\sum_{n=0}^\infty(\frac{1}{3^n})\)

\(=5\sum_{n=0}^\infty(\frac{1}{2^n})-\sum_{n=0}^\infty(\frac{1}{3^n})\)

\(=5(\frac{\frac12}{1-\frac12})-(\frac{\frac13}{1-\frac13})\)

\(=5(\frac{\frac12}{\frac12})-(\frac{\frac13}{\frac23})\)

\(=5(1)-(\frac12)\)

\(=\frac92\)