The sum of the series Σ (n=1 to ∞) (n²+6n+10)/(2n+1)! is equal to : (1) 41/8 e + 19/8 e⁻¹ - 10 (2) -

Q: Wait, why do we change $n$ to $r=2n+1$? Can't we just solve it with $n$?

You could, but it would be a nightmare! The factorial $(2n+1)!$ is difficult to manipulate directly because the index $n$ is inside the factorial. By substituting $r=2n+1$, we transform the denominator into a simple $r!$, which is the standard form for Taylor series expansions. It is all about making the algebra work for you, not against you.

Q: How do I know which terms to split the numerator into? Is there a trick?

Yes, the trick is to look at the denominator. Since you have $r!$ in the denominator, you want to create terms in the numerator that can cancel out factors of $r!$. Specifically, $r(r-1)$ cancels with $r!$ to leave $1/(r-2)!$, and $r$ cancels to leave $1/(r-1)!$. By decomposing the numerator into $r(r-1)$, $r$, and a constant, you are essentially forcing the series into a form that matches the Taylor series for $e^x$.

Q: Why did the summation range change to odd numbers?

Great catch! The original series is $\sum_{n=1}^{\infty}$. When we define $r = 2n+1$, we must map the values of $n$ to $r$. For $n=1$, $r=3$. For $n=2$, $r=5$. For $n=3$, $r=7$. Thus, $r$ takes all odd integer values starting from $3$. If you sum over all $r$, you will get the wrong answer because you would be including even values of $r$ that were never in the original series.

Q: What if I forgot the expansion of $e$ and $e^{-1}$?

Don't panic! Remember that $e = \sum_{k=0}^{\infty} \frac{1}{k!}$ and $e^{-1} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!}$. By adding these two, you cancel out the odd terms, leaving twice the sum of even terms. By subtracting them, you cancel out the even terms, leaving twice the sum of odd terms. This is a fundamental tool in your JEE arsenal—keep it handy!

Question

The sum of the series $\sum_{n=1}^{\infty} \frac{n^2+6n+10}{(2n+1)!}$ is equal to :

Options:

(A) $\frac{41}{8}e + \frac{19}{8}e^{-1} - 10$

(B) $-\frac{41}{8}e + \frac{19}{8}e^{-1} - 10$

(C) $\frac{41}{8}e - \frac{19}{8}e^{-1} - 10$

(D) $\frac{41}{8}e + \frac{19}{8}e^{-1} + 10$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The series is given by $\sum_{n=1}^{\infty} \frac{n^2+6n+10}{(2n+1)!}$. At first glance, the quadratic numerator over a factorial denominator suggests an exponential series structure.

The presence of $(2n+1)!$ is a significant hint. We recall the standard Taylor series for $e^x$ and $e^{-x}$:

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

### The Factorial Trap

To simplify the expression, we perform a substitution. Let $r = 2n+1$. As $n$ ranges from $1$ to $\infty$, $r$ takes the values $3, 5, 7, \dots$ (all odd integers starting from $3$).

From $r = 2n+1$, we have $n = \frac{r-1}{2}$. Substituting this into the numerator $n^2+6n+10$:

$$\left( \frac{r-1}{2} ight)^2 + 6\left( \frac{r-1}{2} ight) + 10 = \frac{r^2-2r+1}{4} + 3(r-1) + 10$$

Simplifying the expression with a common denominator of $4$:

$$\frac{r^2-2r+1 + 12r - 12 + 40}{4} = \frac{r^2+10r+29}{4}$$

### The Art of Decomposition

We now express the general term $T_r$ as:

$$T_r = \frac{r^2+10r+29}{4 \cdot r!}$$

To facilitate summation, we decompose the numerator $r^2+10r+29$ into terms that allow cancellation with $r!$:

$$r^2+10r+29 = r(r-1) + 11r + 29$$

Substituting this back into $T_r$:

$$T_r = \frac{1}{4} \left( \frac{r(r-1)}{r!} + \frac{11r}{r!} + \frac{29}{r!} ight) = \frac{1}{4} \left( \frac{1}{(r-2)!} + \frac{11}{(r-1)!} + \frac{29}{r!} ight)$$

### The Summation Dance

We sum over odd $r \in \{3, 5, 7, \dots\}$. We utilize the known series for odd and even positions:
Sum of odd factorials: $\sum_{k 	ext{ odd}} \frac{1}{k!} = \frac{e-e^{-1}}{2}$
Sum of even factorials: $\sum_{k 	ext{ even}} \frac{1}{k!} = \frac{e+e^{-1}}{2}$

1. For $\sum \frac{1}{(r-2)!}$ where $r=3, 5, \dots$: This is $\frac{1}{1!} + \frac{1}{3!} + \dots = \frac{e-e^{-1}}{2}$.
2. For $\sum \frac{1}{(r-1)!}$ where $r=3, 5, \dots$: This is $\frac{1}{2!} + \frac{1}{4!} + \dots = \left( \frac{e+e^{-1}}{2} - \frac{1}{0!} ight) = \frac{e+e^{-1}-2}{2}$.
3. For $\sum \frac{1}{r!}$ where $r=3, 5, \dots$: This is $\frac{1}{3!} + \frac{1}{5!} + \dots = \left( \frac{e-e^{-1}}{2} - \frac{1}{1!} ight) = \frac{e-e^{-1}-2}{2}$.

### Final Calculation

Combining these results into the sum $S$:

$$S = \frac{1}{4} \left[ \frac{e-e^{-1}}{2} + 11 \left( \frac{e+e^{-1}-2}{2} ight) + 29 \left( \frac{e-e^{-1}-2}{2} ight) ight]$$

Factoring out $\frac{1}{2}$:

$$S = \frac{1}{8} \left[ (1+11+29)e + (-1+11-29)e^{-1} - 22 - 58 ight]$$

$$S = \frac{1}{8} [ 41e - 19e^{-1} - 80 ]$$

The final result is:

$$\mathbf{S = \frac{41}{8}e - \frac{19}{8}e^{-1} - 10}$$

Animated Solution for Mathematics - Sequence and Series: The sum of the series $\sum_{n = 1}^{\infty} \frac{n ^{2} + 6 n + 10}{( 2 n + 1 )!}$ is equal to :

Select Answer:

Visualized Solution

Introduction to the Series

Substitution for Simplification

Transforming the Numerator

Simplifying the Numerator Expression

Defining the General Term $T_{r}$

Decomposing the Numerator

Splitting into Partial Terms

Determining the Summation Range

Recalling the Exponential Series

Evaluating the First Sum

Evaluating the Second Sum

Evaluating the Third Sum

Substituting Series Values

Final Algebraic Calculation

Conclusion and Key Takeaways

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Factorial Trap

The Art of Decomposition

The Summation Dance

Final Calculation

Similar Questions

The sum of series $\frac{1}{2 !} + \frac{1}{4 !} + \frac{1}{6 !} + \dots$ is

The sum of series $\frac{1}{2 !} - \frac{1}{3 !} + \frac{1}{4 !} - \dots$ upto infinity is

The sum of the series $1 + \frac{1}{4.2 !} + \frac{1}{16.4 !} + \frac{1}{64.6 !} + \dots$ ad inf. is

The value of $\sum_{k = 1}^{\infty} (- 1)^{k + 1} \frac{k ( k + 1 )}{k !}$ is

The sum of the series $\frac{1}{1 - 3 \cdot 1 ^{2} + 1 ^{4}} + \frac{2}{1 - 3 \cdot 2 ^{2} + 2 ^{4}} + \frac{3}{1 - 3 \cdot 3 ^{2} + 3 ^{4}} + \dots$ up to 10 terms is

The sum to 10 terms of the series $\frac{1}{1 + 1 ^{2} + 1 ^{4}} + \frac{2}{1 + 2 ^{2} + 2 ^{4}} + \frac{3}{1 + 3 ^{2} + 3 ^{4}} + \dots$ is:-

Let $S_{n} = 1 \cdot (n - 1) + 2 \cdot (n - 2) + 3 \cdot (n - 3) + \dots + (n - 1) \cdot 1, n \geq 4$ . The sum $\sum_{n = 4}^{\infty} (\frac{2 S _{n}}{n !} - \frac{1}{( n - 2 )!})$ is equal to :

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

The sum of $10$ terms of the series $\frac{3}{1 ^{2} \times 2 ^{2}} + \frac{5}{2 ^{2} \times 3 ^{2}} + \frac{7}{3 ^{2} \times 4 ^{2}} + \dots$ is :

The value of $lim_{n \to \infty} (\sum_{k = 1}^{n} \frac{k ^{3} + 6 k ^{2} + 11 k + 5}{( k + 3 )!})$ is equal to:

Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the series ∑n=1∞​(2n+1)!n2+6n+10​ is equal to :