Answer:
207
Step-by-step explanation:
[tex]f(n)=log_34\cdot log_45\cdot log_56\cdot ...\cdot log_{n-1}n[/tex]
[tex]=\frac{log4}{log3} \cdot\frac{log5}{log4} \cdot\frac{log6}{log5} \cdot ...\cdot\frac{log(n)}{log(n-1)}[/tex]
[tex]=\frac{log(n)}{log3}[/tex]
[tex]\sum\limits_{k=3}^{20} f(3^k)=f(3^3)+f(3^4)+f(3^5)+...+f(3^{20})[/tex]
[tex]=\frac{log3^3}{log3} +\frac{log3^4}{log3} +\frac{log3^5}{log3} +...+\frac{log3^{20}}{log3}[/tex]
[tex]=\frac{3(log3)}{log3} +\frac{4(log3)}{log3} +\frac{5(log3)}{log3} +...+\frac{20(log3)}{log3}[/tex]
[tex]=3+4+5+...+20[/tex]
[tex]=\frac{18}{2} (3+20)[/tex] → arithmetic series
[tex]=207[/tex]
