Animated Solution for Mathematics - Sequence and Series: Let ${\sum }_{n=0}^{\infty }\frac{n^3((2n)!)+(2n-1)(n!)}{(n!)((2n)!)}=ae+\frac{b}{e}+c$, where $a,b,c\in \mathbb{Z}$ and $e={\sum }_{n=0}^{\infty }\frac{1}{n!}$. Then $a^2-b+c$ is equal to

Imagine you are standing on the precipice of a complex series problem. You look at the expression:

It looks like a chaotic mess of factorials, but in the world of JEE Advanced, chaos is often just order in disguise. Our goal is to tame this beast and reduce it to the form $ae+\frac{b}{e}+c$.

The first rule of mathematical warfare is to simplify the battlefield. We cannot handle the expression as a single, monolithic fraction. Instead, we use the linearity of summation to split the numerator over the common denominator:

Notice the beauty of this step. In the first term, the $(2n)!$ cancels out. In the second term, the $n!$ vanishes. We are left with two distinct, manageable series, which we shall call $S_1$ and $S_2$.

Now, let us focus on $S_1={\sum }_{n=0}^{\infty }\frac{n^3}{n!}$. You might be tempted to try and cancel $n^3$ with $n!$, but that is a trap. Instead, we use the 'falling factorial' trick. We express $n^3$ as a combination of terms that match the denominator:

Why do we do this? Because when we divide by $n!$, the terms cancel perfectly:

Substituting these back, $S_1$ becomes a sum of three series, each of which is the Taylor expansion of $e$. Thus, $S_1=e+3e+e=5e$.

Now for $S_2={\sum }_{n=0}^{\infty }\frac{2n-1}{(2n)!}$. We split this again: $\sum \frac{2n}{(2n)!}-\sum \frac{1}{(2n)!}$. The first part simplifies to $\sum \frac{1}{(2n-1)!}$.

We know that $e=\sum \frac{1}{n!}$ and $e^{-1}=\sum \frac{(-1{)}^n}{n!}$. By manipulating these, we find that the sum of odd factorials is $\frac{e-e^{-1}}{2}$ and the sum of even factorials is $\frac{e+e^{-1}}{2}$.

When we perform the subtraction for $S_2$, the $e$ terms cancel out, leaving us with $-e^{-1}$, or $-\frac{1}{e}$.

We have arrived at the finish line. Our total sum is $S=5e-\frac{1}{e}$. Comparing this to our target form $ae+\frac{b}{e}+c$, we identify $a=5$, $b=-1$, and $c=0$.

The final calculation, $a^2-b+c$, becomes:

You have just conquered a problem that would make most students tremble. Remember, in mathematics, the complexity is often just a veil; with the right tools, you can always see the truth underneath.