If $$\sum_{n=1}^{5}\frac{1}{n(n+1)(n+2)(n+3)} = \frac{k}{3}$$, then $$k$$ is equal to:

We are given the sum $$\sum_{n=1}^{5}\frac{1}{n(n+1)(n+2)(n+3)} = \frac{k}{3}$$ and need to find the value of $$k$$. To solve this, we will decompose the general term $$\frac{1}{n(n+1)(n+2)(n+3)}$$ using partial fractions.

Assume $$\frac{1}{n(n+1)(n+2)(n+3)} = \frac{a}{n} + \frac{b}{n+1} + \frac{c}{n+2} + \frac{d}{n+3}$$. Multiplying both sides by $$n(n+1)(n+2)(n+3)$$ gives:

$$1 = a(n+1)(n+2)(n+3) + b(n)(n+2)(n+3) + c(n)(n+1)(n+3) + d(n)(n+1)(n+2)$$

We solve for $$a$$, $$b$$, $$c$$, and $$d$$ by substituting convenient values of $$n$$. Setting $$n = 0$$:

$$1 = a(1)(2)(3) + b(0) + c(0) + d(0) = 6a \implies a = \frac{1}{6}$$

Setting $$n = -1$$:

$$1 = a(0) + b(-1)(1)(2) + c(0) + d(0) = -2b \implies b = -\frac{1}{2}$$

Setting $$n = -2$$:

$$1 = a(0) + b(0) + c(-2)(-1)(1) + d(0) = 2c \implies c = \frac{1}{2}$$

Setting $$n = -3$$:

$$1 = a(0) + b(0) + c(0) + d(-3)(-2)(-1) = -6d \implies d = -\frac{1}{6}$$

Thus, the partial fraction decomposition is:

$$\frac{1}{n(n+1)(n+2)(n+3)} = \frac{1/6}{n} - \frac{1/2}{n+1} + \frac{1/2}{n+2} - \frac{1/6}{n+3} = \frac{1}{6}\left(\frac{1}{n} - \frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right)$$

Now, the sum from $$n=1$$ to $$n=5$$ is:

$$\sum_{n=1}^{5} \frac{1}{n(n+1)(n+2)(n+3)} = \frac{1}{6} \sum_{n=1}^{5} \left( \frac{1}{n} - \frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3} \right)$$

Expanding the sum inside:

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

$$\sum_{n=1}^{5} \frac{1}{n+1} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}$$

$$\sum_{n=1}^{5} \frac{1}{n+2} = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}$$

$$\sum_{n=1}^{5} \frac{1}{n+3} = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}$$

Substituting these into the expression:

$$\frac{1}{6} \left[ \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \right) - 3\left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \right) + 3\left( \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \right) - \left( \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) \right]$$

Combining the coefficients for each fraction:

For $$\frac{1}{1}$$: coefficient = 1

For $$\frac{1}{2}$$: $$1$$ (from first sum) $$- 3 \times 1$$ (from second sum) = $$1 - 3 = -2$$

For $$\frac{1}{3}$$: $$1$$ (first) $$- 3 \times 1$$ (second) $$+ 3 \times 1$$ (third) = $$1 - 3 + 3 = 1$$

For $$\frac{1}{4}$$: $$1$$ (first) $$- 3 \times 1$$ (second) $$+ 3 \times 1$$ (third) $$- 1 \times 1$$ (fourth) = $$1 - 3 + 3 - 1 = 0$$

For $$\frac{1}{5}$$: $$1$$ (first) $$- 3 \times 1$$ (second) $$+ 3 \times 1$$ (third) $$- 1 \times 1$$ (fourth) = $$1 - 3 + 3 - 1 = 0$$

For $$\frac{1}{6}$$: $$- 3 \times 1$$ (second) $$+ 3 \times 1$$ (third) $$- 1 \times 1$$ (fourth) = $$-3 + 3 - 1 = -1$$

For $$\frac{1}{7}$$: $$+ 3 \times 1$$ (third) $$- 1 \times 1$$ (fourth) = $$3 - 1 = 2$$

For $$\frac{1}{8}$$: $$- 1 \times 1$$ (fourth) = $$-1$$

So the expression simplifies to:

$$\frac{1}{6} \left[ 1 \cdot \frac{1}{1} + (-2) \cdot \frac{1}{2} + 1 \cdot \frac{1}{3} + 0 \cdot \frac{1}{4} + 0 \cdot \frac{1}{5} + (-1) \cdot \frac{1}{6} + 2 \cdot \frac{1}{7} + (-1) \cdot \frac{1}{8} \right] = \frac{1}{6} \left[ 1 - 2 \cdot \frac{1}{2} + \frac{1}{3} - \frac{1}{6} + 2 \cdot \frac{1}{7} - \frac{1}{8} \right]$$

Simplifying inside the brackets:

$$1 - 2 \cdot \frac{1}{2} = 1 - 1 = 0$$

So we have:

$$\frac{1}{6} \left[ \frac{1}{3} - \frac{1}{6} + \frac{2}{7} - \frac{1}{8} \right]$$

Computing $$\frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6}$$, so:

$$\frac{1}{6} \left[ \frac{1}{6} + \frac{2}{7} - \frac{1}{8} \right]$$

Now, find a common denominator for $$\frac{1}{6}$$, $$\frac{2}{7}$$, and $$\frac{1}{8}$$. The least common multiple of 6, 7, and 8 is $$2^4 \times 3 \times 7 = 336$$.

Converting:

$$\frac{1}{6} = \frac{56}{336}, \quad \frac{2}{7} = \frac{96}{336}, \quad \frac{1}{8} = \frac{42}{336}$$

So:

$$\frac{1}{6} + \frac{2}{7} - \frac{1}{8} = \frac{56}{336} + \frac{96}{336} - \frac{42}{336} = \frac{56 + 96 - 42}{336} = \frac{110}{336}$$

Simplifying $$\frac{110}{336}$$ by dividing numerator and denominator by 2:

$$\frac{110 \div 2}{336 \div 2} = \frac{55}{168}$$

Thus, the expression inside the brackets is $$\frac{55}{168}$$, and the sum is:

$$\frac{1}{6} \times \frac{55}{168} = \frac{55}{1008}$$

Now, the sum equals $$\frac{k}{3}$$, so:

$$\frac{55}{1008} = \frac{k}{3}$$

Solving for $$k$$:

$$k = 3 \times \frac{55}{1008} = \frac{165}{1008}$$

Simplifying $$\frac{165}{1008}$$ by dividing numerator and denominator by 3:

$$\frac{165 \div 3}{1008 \div 3} = \frac{55}{336}$$

Comparing with the options, $$\frac{55}{336}$$ matches option A.

Hence, the correct answer is Option A.