JEE 2023 — Mathematics PYQ

To solve this, we use the standard Taylor series expansions for $e$ and $e^{-1}$:

• $e={\sum }_{n=0}^{\infty }\frac{1}{n!}=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots$

• $e^{-1}={\sum }_{n=0}^{\infty }\frac{(-1{)}^n}{n!}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\dots$

From these, we derive the sums for even factorials:

• $\frac{e+e^{-1}}{2}={\sum }_{n=0}^{\infty }\frac{1}{(2n)!}=1+\frac{1}{2!}+\frac{1}{4!}+\dots$

• $\frac{e-e^{-1}}{2}={\sum }_{n=0}^{\infty }\frac{1}{(2n+1)!}=\frac{1}{1!}+\frac{1}{3!}+\dots$

1. General Term Decomposition

Let the general term be $T_n=\frac{2n^2+3n+4}{(2n)!}$. We need to express the numerator in terms of $(2n)$ to cancel terms in the factorial:

2. Splitting the Sum

Now, substitute this back into the summation:

Simplify each term:

• First term: ${\sum }_{n=1}^{\infty }\frac{1}{2}\cdot \frac{1}{(2n-2)!}$. Let $k=n-1$, then $\frac{1}{2}{\sum }_{k=0}^{\infty }\frac{1}{(2k)!}=\frac{1}{2}\left(\frac{e+e^{-1}}{2}\right)$

• Second term: ${\sum }_{n=1}^{\infty }\frac{2}{(2n-1)!}$. This is $2\left(\frac{e-e^{-1}}{2}\right)=e-e^{-1}$

• Third term: $4{\sum }_{n=1}^{\infty }\frac{1}{(2n)!}$. Note the sum starts at $n=1$, so we must subtract the $n=0$ term (which is 1): $4\left(\frac{e+e^{-1}}{2}-1\right)$

3. Combining the Results

Group the $e$ and $e^{-1}$ terms:

• $e$ terms: $(\frac{1}{4}+1+2)e=\frac{13}{4}e$

• $e^{-1}$ terms: $(\frac{1}{4}-1+2)e^{-1}=\frac{5}{4}{e}^{-1}$

Therefore:

Correct Option: (4)