Given:

\(f(x)=\sin5x\)

From the table of power series, the Maclaurin series for the above function using the table of power series.

\(\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+...=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}\)

Replacing x by 5x,

\(\sin5x=5x-\frac{(5x)^3}{3!}+\frac{(5x)^5}{5!}+\frac{(5x)^7}{7!}+...\)

\(\sin5x=5x-\frac{125x^3}{6}+\frac{325x^5}{24}+\frac{15625x^7}{1008}+...\)

\(\sin5x=\sum_{n=0}^\infty\frac{(-1)^n(5x)^{2n+1}}{(2n+1)!}\)

Hence, \(\sin5x=\sum_{n=0}^\infty\frac{(-1)^n(5x)^{2n+1}}{(2n+1)!}\)

\(f(x)=\sin5x\)

From the table of power series, the Maclaurin series for the above function using the table of power series.

\(\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+...=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}\)

Replacing x by 5x,

\(\sin5x=5x-\frac{(5x)^3}{3!}+\frac{(5x)^5}{5!}+\frac{(5x)^7}{7!}+...\)

\(\sin5x=5x-\frac{125x^3}{6}+\frac{325x^5}{24}+\frac{15625x^7}{1008}+...\)

\(\sin5x=\sum_{n=0}^\infty\frac{(-1)^n(5x)^{2n+1}}{(2n+1)!}\)

Hence, \(\sin5x=\sum_{n=0}^\infty\frac{(-1)^n(5x)^{2n+1}}{(2n+1)!}\)