Answer:
Infinite sequence with terms [tex]t_{n+1}=t_n\cdot ab[/tex] or [tex]t_{n+1}=2a\cdot (ab)^n=2a^{n+1}b^n.[/tex]
[tex]t_7=2a^7b^6.[/tex]
Step-by-step explanation:
Consider the pattern
[tex]2a,\ 2a^2b,\ 2a^3b^2,\ 2a^4b^3,\ ...[/tex]
First term of this pattern is
[tex]t_1=2a,[/tex]
second -
[tex]t_2=t_1\cdot ab,[/tex]
third -
[tex]t_3=t_2\cdot ab,[/tex]
fourth -
[tex]t_4=t_3\cdot ab,[/tex]
fifth -
[tex]t_5=t_4\cdot ab=2a^4b^3\cdot ab=2a^5b^4,[/tex]
sixth -
[tex]t_6=t_5\cdot ab=2a^5b^4\cdot ab=2a^6b^5,[/tex]
seventh -
[tex]t_7=t_6\cdot ab=2a^6b^5\cdot ab=2a^7b^6.[/tex]
This equence is infinite and each next term is obtained from the previous multiplying by ab.