The $$n^{th}$$ terms of two sequences are $$a_n=\frac{3n+1}{2^{n-1}}$$ and $$b_n=\frac{3n-2}{2^n}$$ $$\forall n\in\mathbb{N}$$. $$A_n$$ and $$B_n$$ are the sums of the first $$n$$ terms, respectively, of these sequences. What is the value of $$A_6-B_6$$?

To find the value of $$A_6 - B_6$$, we first analyze the general terms of the two given sequences, $$a_n$$ and $$b_n$$.

The $$n^{\text{th}}$$ terms are given by:

$$a_n = \frac{3n + 1}{2^{n-1}}$$

$$b_n = \frac{3n - 2}{2^n}$$

Since $$A_n$$ and $$B_n$$ represent the sums of the first $$n$$ terms of their respective sequences, the difference $$A_n - B_n$$ can be expressed as the sum of the differences of their general terms:

$$A_n - B_n = \sum_{k=1}^{n} (a_k - b_k)$$

Let us simplify the general term $$a_k - b_k$$:

$$a_k - b_k = \frac{3k + 1}{2^{k-1}} - \frac{3k - 2}{2^k}$$

Taking $$2^k$$ as the common denominator:

$$a_k - b_k = \frac{2(3k + 1) - (3k - 2)}{2^k}$$

$$a_k - b_k = \frac{6k + 2 - 3k + 2}{2^k} = \frac{3k + 4}{2^k}$$

Thus, the sum $$S_n = A_n - B_n$$ becomes an arithmetico-geometric progression (AGP):

$$S_n = \sum_{k=1}^{n} \frac{3k + 4}{2^k} = \frac{7}{2^1} + \frac{10}{2^2} + \frac{13}{2^3} + \dots + \frac{3n + 4}{2^n}$$

To sum this AGP, multiply both sides by the common ratio $$\frac{1}{2}$$:

$$\frac{1}{2} S_n = \frac{7}{2^2} + \frac{10}{2^3} + \dots + \frac{3n - 2}{2^n} + \frac{3n + 4}{2^{n+1}}$$

Subtracting the second equation from the first:

$$\frac{1}{2} S_n = \frac{7}{2} + \left( \frac{3}{2^2} + \frac{3}{2^3} + \dots + \frac{3}{2^n} \right) - \frac{3n + 4}{2^{n+1}}$$

The expression inside the parentheses is a finite geometric series with $$n-1$$ terms, a first term of $$\frac{3}{4}$$, and a common ratio of $$\frac{1}{2}$$:

$$\frac{3}{2^2} + \frac{3}{2^3} + \dots + \frac{3}{2^n} = \frac{\frac{3}{4} \left(1 - \frac{1}{2^{n-1}}\right)}{1 - \frac{1}{2}} = \frac{3}{2} \left(1 - \frac{1}{2^{n-1}}\right)$$

Substituting this back into our expression for $$\frac{1}{2} S_n$$:

$$\frac{1}{2} S_n = \frac{7}{2} + \frac{3}{2} - \frac{3}{2^n} - \frac{3n + 4}{2^{n+1}}$$

$$\frac{1}{2} S_n = 5 - \frac{3}{2^n} - \frac{3n + 4}{2^{n+1}}$$

Multiplying the entire equation by 2 gives the general sum $$S_n$$:

$$S_n = 10 - \frac{6}{2^n} - \frac{3n + 4}{2^n} = 10 - \frac{3n + 10}{2^n}$$

Now, substituting $$n = 6$$ to find $$A_6 - B_6$$:

$$S_6 = 10 - \frac{3(6) + 10}{2^6}$$

$$S_6 = 10 - \frac{18 + 10}{64}$$

$$S_6 = 10 - \frac{28}{64}$$

Simplifying the fraction by dividing the numerator and denominator by 4:

$$S_6 = 10 - \frac{7}{16} = \frac{160 - 7}{16} = \frac{153}{16}$$

The correct option is B.