Let $$a_1 = b_1 = 1$$ and $$a_n = a_{n-1} + (n-1)$$, $$b_n = b_{n-1} + a_{n-1}$$, $$\forall n \geq 2$$. If $$S = \sum_{n=1}^{10} \left(\frac{b_n}{2^n}\right)$$ and $$T = \sum_{n=1}^{8} \frac{n}{2^{n-1}}$$ then $$2^7(2S - T)$$ is equal to

Compute $$2^7(2S - T)$$ where $$S = \sum_{n=1}^{10} \frac{b_n}{2^n}$$ and $$T = \sum_{n=1}^{8} \frac{n}{2^{n-1}}$$.

Given $$a_1 = b_1 = 1$$, $$a_n = a_{n-1} + (n-1)$$, $$b_n = b_{n-1} + a_{n-1}$$.

$$a_n = 1 + \frac{n(n-1)}{2}$$: values are $$1, 2, 4, 7, 11, 16, 22, 29, 37, 46$$

$$b_n$$: values are $$1, 2, 4, 8, 15, 26, 42, 64, 93, 130$$

$$S = \frac{1}{2} + \frac{2}{4} + \frac{4}{8} + \frac{8}{16} + \frac{15}{32} + \frac{26}{64} + \frac{42}{128} + \frac{64}{256} + \frac{93}{512} + \frac{130}{1024}$$

Converting to a common denominator of 1024:

$$S = \frac{512 + 512 + 512 + 512 + 480 + 416 + 336 + 256 + 186 + 130}{1024} = \frac{3852}{1024} = \frac{963}{256}$$

$$T = 1 + 1 + \frac{3}{4} + \frac{4}{8} + \frac{5}{16} + \frac{6}{32} + \frac{7}{64} + \frac{8}{128}$$

Converting to a common denominator of 128:

$$T = \frac{128 + 128 + 96 + 64 + 40 + 24 + 14 + 8}{128} = \frac{502}{128} = \frac{251}{64}$$

$$2S = \frac{1926}{256} = \frac{963}{128}$$

$$2S - T = \frac{963}{128} - \frac{251}{64} = \frac{963}{128} - \frac{502}{128} = \frac{461}{128}$$

$$2^7(2S - T) = 128 \times \frac{461}{128} = 461$$

The answer is $$\boxed{461}$$.