Provide the first six terms of the following sequence:

t1=81 tn=1/3tn-1

We get tne n-th term with: tn=1/3tn-1

We know that the first term is t1=81, so 81=1/3t*1-1

81=1/3t-1

82=1/3t

82*3t=1

3t=1/82

t=1/82*3=1/246

The second term is: 1/

2 * (1/246) * 2) - 1

We get the third term by replacing n with 3 and so on ...