1. Identify the general term

The series is

$$1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\cdots$$

Notice that in the denominator we have $(n+1)!$ when the numerator is the sum of the first $n$ odd numbers.

We use the fact:

$$1+3+5+\cdots +(2n-1)=n^2$$

So the general term is

$$\frac{n^2}{(n+1)!}, \qquad n=1,2,3,\dots$$

Thus,

$$S=\sum_{n=1}^{\infty} \frac{n^2}{(n+1)!}$$

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2. Rewrite the numerator

We simplify $n^2$ in a form useful with factorials:

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

So,

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

Now simplify each part:

$$\frac{n(n-1)}{(n+1)!}=\frac{1}{(n+1)(n-2)!}\quad (n\ge 2)$$

A better decomposition is obtained by expressing $n^2$ as:

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

Hence,

$$\frac{n^2}{(n+1)!}=\frac{(n+1)n}{(n+1)!}-\frac{n}{(n+1)!}$$ $$=\frac{1}{(n-1)!}-\frac{1}{n!}+\frac{1}{(n+1)!}?$$

Let us verify carefully.

Since

$$\frac{n}{(n+1)!}=\frac{(n+1)-1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$$

Then

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

But this is not the cleanest path.

Let us use a standard decomposition:

$$n^2=(n+1)n-n=(n+1)n-[(n+1)-1]=n(n+1)-(n+1)+1$$

So,

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

Now,

$$\frac{n-1}{n!}=\frac{n}{n!}-\frac{1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}$$

Hence,

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

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3. Sum the series

Therefore,

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

Split the sum:

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

Now evaluate each:

• First sum:

$$\sum_{n=1}^{\infty}\frac{1}{(n-1)!}=\sum_{k=0}^{\infty}\frac{1}{k!}=e$$

• Second sum:

$$\sum_{n=1}^{\infty}\frac{1}{n!}=e-1$$

• Third sum: Let $m=n+1$, then $m=2,3,4,\dots$

$$\sum_{n=1}^{\infty}\frac{1}{(n+1)!}=\sum_{m=2}^{\infty}\frac{1}{m!}=e-\left(1+1\right)=e-2$$

So,

$$S=e-(e-1)+(e-2)=e+1+e-2= e-1+e = 2e-1?$$

Let us simplify correctly:

$$S=e-(e-1)+(e-2)=e-e+1+e-2=e-1$$

This is not among the options, so we should re-check the decomposition/indexing.

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4. Re-evaluate the series carefully

The given terms are:

$$1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\cdots$$

Now:

• First term $1 = \dfrac{1^2}{1!}$
• Second term $\dfrac{1+3}{2!}=\dfrac{2^2}{2!}$
• Third term $\dfrac{1+3+5}{3!}=\dfrac{3^2}{3!}$
• Fourth term $\dfrac{1+3+5+7}{4!}=\dfrac{4^2}{4!}$

So the correct general term is actually

$$\frac{n^2}{n!}, \qquad n=1,2,3,\dots$$

Hence,

$$S=\sum_{n=1}^{\infty}\frac{n^2}{n!}$$

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5. Simplify the correct general term

Use

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

Thus,

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

Therefore,

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

Interpret carefully:

• For the first sum, $\frac{1}{(n-2)!}$ starts effectively from $n=2$.

So,

$$\sum_{n=1}^{\infty}\frac{n(n-1)}{n!}=\sum_{n=2}^{\infty}\frac{1}{(n-2)!}=\sum_{k=0}^{\infty}\frac{1}{k!}=e$$

And,

$$\sum_{n=1}^{\infty}\frac{n}{n!}=\sum_{n=1}^{\infty}\frac{1}{(n-1)!}=\sum_{k=0}^{\infty}\frac{1}{k!}=e$$

Hence,

$$S=e+e=2e$$

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6. Match with options

$$S=2e$$

So the correct option is:

B: $2e$

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7. Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

They agree.