The sum $$1 + \dfrac{1 + 3}{2!} + \dfrac{1 + 3 + 5}{3!} + \dfrac{1 + 3 + 5 + 7}{4!} + \ldots$$ upto $$\infty$$ terms, is equal to

The general term of the series is $$T_n = \dfrac{1 + 3 + 5 + \ldots + (2n-1)}{n!}$$.

The sum of first $$n$$ odd numbers is $$n^2$$. So $$T_n = \dfrac{n^2}{n!}$$.

The required sum is $$S = \displaystyle\sum_{n=1}^{\infty} \dfrac{n^2}{n!}$$.

We write $$n^2 = n(n-1) + n$$, so:

$$S = \displaystyle\sum_{n=1}^{\infty} \dfrac{n(n-1)}{n!} + \displaystyle\sum_{n=1}^{\infty} \dfrac{n}{n!}$$

$$= \displaystyle\sum_{n=2}^{\infty} \dfrac{1}{(n-2)!} + \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{(n-1)!}$$

$$= \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{k!} + \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{k!} = e + e = 2e$$

Hence, the correct answer is Option D.