If $$\frac{1}{2 \times 3 \times 4} + \frac{1}{3 \times 4 \times 5} + \frac{1}{4 \times 5 \times 6} + \ldots + \frac{1}{100 \times 101 \times 102} = \frac{k}{101}$$, then $$34k$$ is equal to _______

We need to evaluate $$\frac{1}{2 \times 3 \times 4} + \frac{1}{3 \times 4 \times 5} + \cdots + \frac{1}{100 \times 101 \times 102} = \frac{k}{101}$$ and find $$34k$$.

We use the telescoping identity $$\frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left(\frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)}\right)$$, which can be verified by noting that $$\frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)} = \frac{(r+2) - r}{r(r+1)(r+2)} = \frac{2}{r(r+1)(r+2)}$$.

Applying this to each term of the sum with $$r$$ running from 2 to 100, we get a telescoping series:

$$\sum_{r=2}^{100}\frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left(\frac{1}{2 \cdot 3} - \frac{1}{101 \cdot 102}\right) = \frac{1}{2}\left(\frac{1}{6} - \frac{1}{10302}\right)$$

We bring these fractions to a common denominator. Since $$10302 = 6 \times 1717$$, we write $$\frac{1}{6} = \frac{1717}{10302}$$, and so:

$$\frac{1}{2} \cdot \frac{1717 - 1}{10302} = \frac{1716}{2 \times 10302} = \frac{858}{10302}$$

Now $$10302 = 101 \times 102$$, so this equals $$\frac{858}{101 \times 102}$$. Setting this equal to $$\frac{k}{101}$$, we get $$k = \frac{858}{102} = \frac{143}{17}$$.

Therefore $$34k = 34 \times \frac{143}{17} = 2 \times 143 = 286$$.

Hence, the correct answer is 286.