Find the sum of the first $\mathrm{n}$ terms of the indicated geometric sequence with the given values. \[ \frac{1}{6}+\frac{1}{2}+\frac{3}{2}+\cdots+\frac{81}{2} \] The sum is (Simplify your answer.)

Step 1 ：Given the geometric sequence \(\frac{1}{6}, \frac{1}{2}, \frac{3}{2}, \ldots, \frac{81}{2}\), we can see that the common ratio (r) is 3 and the first term (a) is \(\frac{1}{6}\).

Step 2 ：We can find the number of terms (n) by setting the nth term of the geometric sequence equal to the last term and solving for n. The nth term of a geometric sequence is given by \(a_n = ar^{n-1}\).

Step 3 ：Setting this equal to the last term gives us \(\frac{81}{2} = \frac{1}{6} * 3^{n-1}\). Solving this equation gives us n = 6.

Step 4 ：We can now use the formula for the sum of a geometric sequence to find the sum of the first n terms. The formula is \(S_n = \frac{a(1 - r^n)}{1 - r}\).

Step 5 ：Substituting the values we have, we get \(S_n = \frac{\frac{1}{6}(1 - 3^6)}{1 - 3}\).

Step 6 ：Solving this gives us \(S_n = 60.67\) (rounded to two decimal places).

Step 7 ：Final Answer: The sum of the first n terms of the geometric sequence is \(\boxed{60.67}\).