Sum of Consecutive Odd Numbers

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.

1. Find the sum S of the infinite series

The given series is:

$$S = 1 + \frac{1+3}{2!} + \frac{1+3+5}{3!} + \frac{1+3+5+7}{4!} + \dots$$

The numerator of the $$k^{th}$$ term is the sum of the first k odd numbers, which is known to be $$k^2$$. Thus, the general term $$T_k$$ is:

$$T_k = \frac{k^2}{k!}$$

We can simplify $$T_k$$ to make it easier to sum using the Taylor series for $$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$:

1. For $$k \geq 1$$: $$T_k = \frac{k^2}{k!} = \frac{k \cdot k}{k(k-1)!} = \frac{k}{(k-1)!}$$

2. Let $$m = k-1$$: $$T_k = \frac{m+1}{m!} = \frac{m}{m!} + \frac{1}{m!} = \frac{1}{(m-1)!} + \frac{1}{m!}$$

Summing from $$k=1$$ to $$\infty$$:

$$S = \sum_{k=1}^{\infty} \left( \frac{1}{(k-2)!} + \frac{1}{(k-1)!} \right)$$

The first part $$\sum_{k=2}^{\infty} \frac{1}{(k-2)!}$$ is $$\frac{1}{0!} + \frac{1}{1!} + \dots = e$$.

The second part $$\sum_{k=1}^{\infty} \frac{1}{(k-1)!}$$ is $$\frac{1}{0!} + \frac{1}{1!} + \dots = e$$.

Therefore, $$S = e + e = 2e$$.

2. Set up the Arithmetic Progression (AP)

We are given an AP with:

First term (a): $$e$$

Common difference (d): $$2e$$

Sum of first n terms (Sn): $$8S$$

Substitute $$S = 2e$$:

$$S_n = 8(2e) = 16e$$

3. Solve for n

The formula for the sum of the first n terms of an AP is $$S_n = \frac{n}{2} [2a + (n-1)d]$$. Substituting our values:

$$16e = \frac{n}{2} [2(e) + (n-1)(2e)]$$

Divide both sides by e (since $$e \neq 0$$):

$$16 = \frac{n}{2} [2 + 2n - 2]$$

$$16 = \frac{n}{2} [2n]$$

$$16 = n^2$$

$$n = 4$$