The sum of all possible values of $$n\epsilon N$$, so that the coefficients of $$x,x^{2}\text{ and }x^{3}$$ in the expansion of $$(1+x^{2})^{2}(1+x)^{n}$$, are in arithmetic progression is:

We need to find the coefficients of $$x$$, $$x^2$$, and $$x^3$$ in the expansion of $$(1+x^2)^2(1+x)^n$$ and set them in arithmetic progression (AP).

Since $$(1+x^2)^2 = 1 + 2x^2 + x^4$$, we can multiply this by $$(1+x)^n$$ to find the needed coefficients.

We recall that $$(1+x)^n = \sum_{r=0}^{n} \binom{n}{r}x^r$$, so each term in the product arises from multiplying one term in $$(1+x^2)^2$$ by one term in this expansion.

Only the product of $$1$$ from the first factor with $$\binom{n}{1}x$$ from the second contributes to the coefficient of $$x$$, giving $$\binom{n}{1} = n$$.

For the coefficient of $$x^2$$, the contributions come from $$1 \cdot \binom{n}{2}x^2$$ and $$2x^2 \cdot \binom{n}{0}$$, which sum to $$\binom{n}{2} + 2 = \frac{n(n-1)}{2} + 2$$.

To find the coefficient of $$x^3$$, we combine $$1 \cdot \binom{n}{3}x^3$$ with $$2x^2 \cdot \binom{n}{1}x$$, yielding $$\binom{n}{3} + 2n = \frac{n(n-1)(n-2)}{6} + 2n$$.

Because three numbers $$a,b,c$$ are in AP precisely when $$2b = a + c$$, we set

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

which simplifies to

$$n(n-1) + 4 = 3n + \frac{n(n-1)(n-2)}{6}$$

$$n^2 - n + 4 = 3n + \frac{n(n-1)(n-2)}{6}$$

$$n^2 - 4n + 4 = \frac{n(n-1)(n-2)}{6}$$

$$6(n^2 - 4n + 4) = n(n-1)(n-2)$$

$$6n^2 - 24n + 24 = n^3 - 3n^2 + 2n$$

$$n^3 - 9n^2 + 26n - 24 = 0$$

Testing $$n = 2$$ gives $$8 - 36 + 52 - 24 = 0$$, so $$n = 2$$ is a root. Factoring yields $$(n-2)(n^2 - 7n + 12) = 0$$ or equivalently $$(n-2)(n-3)(n-4) = 0$$, giving $$n = 2, 3, 4$$.

All three values are natural numbers, so their sum is $$2 + 3 + 4 = 9$$.

Hence the correct answer is Option 2: 9.