JEE Main 2024 (04 Apr Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Sequence and Series: Let $a = 1 + \frac{^{2} C _{2}}{3 !} + \frac{^{3} C _{2}}{4 !} + \frac{^{4} C _{2}}{5 !} + \dots$ , $b = 1 + \frac{^{1} C _{0} + ^{1} C _{1}}{1 !} + \frac{^{2} C _{0} + ^{2} C _{1} + ^{2} C _{2}}{2 !} + \frac{^{3} C _{0} + ^{3} C _{1} + ^{3} C _{2} + ^{3} C _{3}}{3 !} + \dots$ . Then $\frac{2 b}{a ^{2}}$ is equal to

Q: Wait, why does the sum of binomial coefficients equal $2^n$?

Think of the Binomial Theorem: $(1+x)^n = \sum_{r=0}^{n} {}^nC_r x^r$. If you set $x=1$, the left side becomes $(1+1)^n = 2^n$, and the right side becomes the sum of all binomial coefficients. It is a beautiful, direct consequence of the theorem!

Q: Why did we get $e-2$ for the second sum in series $a$?

Great catch! Remember that the standard expansion for $e$ is $1 + \frac{1}{1!} + \frac{1}{2!} + \dots$. Our sum started at $n=2$, which corresponds to $\frac{1}{2!} + \frac{1}{3!} + \dots$. Since the full series is $e$, we must subtract the first two terms ($1$ and $\frac{1}{1!}$) to get the remaining sum. Hence, $e - 1 - 1 = e - 2$.

Enter Numerical Value:

Visualized Solution

Understanding Series $b$

Given series: $b = 1 + \frac{^{1} C _{0} + ^{1} C _{1}}{1 !} + \frac{^{2} C _{0} + ^{2} C _{1} + ^{2} C _{2}}{2 !} + \dots$
Recall the binomial property: $\sum_{r = 0}^{n}^{n} C_{r} = 2^{n}$

Simplifying Series $b$

Substitute $2^{n}$ into the numerators.
$b = 1 + \frac{2 ^{1}}{1 !} + \frac{2 ^{2}}{2 !} + \frac{2 ^{3}}{3 !} + \dots$

Evaluating Series $b$ using $e^{x}$

The expansion of $e^{x}$ is: $e^{x} = 1 + \frac{x}{1 !} + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots$
Comparing with $b$ , we see $x = 2$ .
Therefore, $b = e^{2}$ .

Analyzing Series $a$

Series $a = 1 + \frac{^{2} C _{2}}{3 !} + \frac{^{3} C _{2}}{4 !} + \frac{^{4} C _{2}}{5 !} + \dots$
We can write this as: $a = 1 + \sum_{n = 2}^{\infty} \frac{^{n} C _{2}}{( n + 1 )!}$

Expanding the General Term $T_{n}$

The general term for $n \geq 2$ is $T_{n} = \frac{^{n} C _{2}}{( n + 1 )!}$
Expand $^{n} C_{2} = \frac{n ( n - 1 )}{2}$
$T_{n} = \frac{n ( n - 1 )}{2 ( n + 1 )!}$

Algebraic Manipulation of the Numerator

Rewrite the numerator: $n (n - 1) = n^{2} - n$
We want terms that relate to $(n + 1)$ .
Add and subtract $2 n$ : $n^{2} - n = (n^{2} + n) - 2 n = n (n + 1) - 2 n$

Splitting $T_{n}$ into Two Fractions

Substitute the rewritten numerator back into $T_{n}$ .
$T_{n} = \frac{n ( n + 1 ) - 2 n}{2 ( n + 1 )!}$
Split into two parts: $T_{n} = \frac{n ( n + 1 )}{2 ( n + 1 )!} - \frac{2 n}{2 ( n + 1 )!}$

Simplifying the First Fraction

First fraction: $\frac{n ( n + 1 )}{2 ( n + 1 )!}$
Expand the factorial: $(n + 1)! = (n + 1) \cdot n \cdot (n - 1)!$
Cancel $n (n + 1)$ : $\frac{n ( n + 1 )}{2 ( n + 1 ) n ( n - 1 )!} = \frac{1}{2 ( n - 1 )!}$

Simplifying the Second Fraction

Second fraction: $\frac{2 n}{2 ( n + 1 )!} = \frac{n}{( n + 1 )!}$
Add and subtract $1$ in the numerator: $n = (n + 1) - 1$
$\frac{( n + 1 ) - 1}{( n + 1 )!} = \frac{n + 1}{( n + 1 )!} - \frac{1}{( n + 1 )!}$

Final Form of the Second Fraction

Simplify $\frac{n + 1}{( n + 1 )!}$
$(n + 1)! = (n + 1) \cdot n!$
$\frac{n + 1}{( n + 1 ) n !} = \frac{1}{n !}$
So, the second part becomes: $\frac{1}{n !} - \frac{1}{( n + 1 )!}$

Combining to get Final $T_{n}$

Combine the simplified parts of $T_{n}$ .
$T_{n} = \frac{1}{2 ( n - 1 )!} - [\frac{1}{n !} - \frac{1}{( n + 1 )!}]$
$T_{n} = \frac{1}{2 ( n - 1 )!} - \frac{1}{n !} + \frac{1}{( n + 1 )!}$

Setting up the Sum for $a$

The total sum is $a = 1 + \sum_{n = 2}^{\infty} T_{n}$
Substitute $T_{n}$ :
$a = 1 + \frac{1}{2} \sum_{n = 2}^{\infty} \frac{1}{( n - 1 )!} - \sum_{n = 2}^{\infty} \frac{1}{n !} + \sum_{n = 2}^{\infty} \frac{1}{( n + 1 )!}$

Evaluating the Infinite Sums

Recall $e = 1 + \frac{1}{1 !} + \frac{1}{2 !} + \frac{1}{3 !} + \dots$
$\sum_{n = 2}^{\infty} \frac{1}{( n - 1 )!} = \frac{1}{1 !} + \frac{1}{2 !} + \dots = e - 1$
$\sum_{n = 2}^{\infty} \frac{1}{n !} = \frac{1}{2 !} + \frac{1}{3 !} + \dots = e - 2$
$\sum_{n = 2}^{\infty} \frac{1}{( n + 1 )!} = \frac{1}{3 !} + \frac{1}{4 !} + \dots = e - \frac{5}{2}$

Calculating the Value of $a$

Substitute the sums back into $a$ :
$a = 1 + \frac{1}{2} (e - 1) - (e - 2) + (e - \frac{5}{2})$
$a = 1 + \frac{e}{2} - \frac{1}{2} - e + 2 + e - \frac{5}{2}$
Grouping $e$ terms: $\frac{e}{2} - e + e = \frac{e}{2}$
Grouping constants: $1 - 0.5 + 2 - 2.5 = 0$
Therefore, $a = \frac{e}{2}$

Final Calculation of $\frac{2 b}{a ^{2}}$

We found $a = \frac{e}{2}$ and $b = e^{2}$ .
We need to evaluate $\frac{2 b}{a ^{2}}$ .
Substitute the values: $\frac{2 ( e ^{2} )}{( \frac{e}{2} ) ^{2}}$
$= \frac{2 e ^{2}}{\frac{e ^{2}}{4}} = 2 e^{2} \cdot \frac{4}{e ^{2}}$
$= 8$

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The Sigma Insight: Sum of Special Series

Analyzing the Setup

Welcome, future engineer! Today, we are not just solving a problem; we are peeling back the layers of a mathematical onion. Infinite series can look intimidating, like a chaotic mess of factorials and binomial coefficients, but beneath that complexity lies a profound, elegant order.

Decoding Series $b$

Imagine you are looking at the series

b = 1 + \frac{^{1} C _{0} + ^{1} C _{1}}{1 !} + \frac{^{2} C _{0} + ^{2} C _{1} + ^{2} C _{2}}{2 !} + \dots

. At first glance, it looks like a nightmare of combinations. But pause and look at the pattern in the numerators.

They are sums of binomial coefficients. Recall the fundamental identity:

\sum_{r = 0}^{n}^{n} C_{r} = 2^{n}

. Suddenly, the fog lifts!

The series transforms into:

b = 1 + \frac{2 ^{1}}{1 !} + \frac{2 ^{2}}{2 !} + \frac{2 ^{3}}{3 !} + \dots

Does this look familiar? It is the Maclaurin expansion of

e^{x}

where

x = 2

. Just like that, $b = e^{2}$ . We have conquered the first mountain.

The Challenge of Series $a$

Now, we face series

a = 1 + \frac{^{2} C _{2}}{3 !} + \frac{^{3} C _{2}}{4 !} + \frac{^{4} C _{2}}{5 !} + \dots

. This is the true test. We define the general term as

T_{n} = \frac{^{n} C _{2}}{( n + 1 )!}

We know that

^{n} C_{2} = \frac{n ( n - 1 )}{2}

. So, the general term becomes:

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

Here is where the 'algebraic surgery' begins. We need to force the numerator to look like the denominator. We rewrite

n (n - 1)

n (n + 1) - 2 n

The Algebraic Surgery

By splitting the fraction, we get:

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

The first part simplifies beautifully:

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

The second part,

\frac{n}{( n + 1 )!}

, requires one more trick: add and subtract

1

in the numerator to get

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

. Now, our series

a

is just a collection of standard exponential series terms.

The Grand Finale

We distribute the summation across these terms. We have

a = 1 + \frac{1}{2} \sum \frac{1}{( n - 1 )!} - \sum \frac{1}{n !} + \sum \frac{1}{( n + 1 )!}

By carefully evaluating these sums starting from

n = 2

, we find:

a = 1 + \frac{1}{2} (e - 1) - (e - 2) + (e - \frac{5}{2})

When the dust settles, the constants cancel, and we are left with $a = \frac{e}{2}$ .

Finally, we calculate the requested value:

\frac{2 b}{a ^{2}} = \frac{2 ( e ^{2} )}{( e /2 ) ^{2}} = \frac{2 e ^{2}}{e ^{2} /4} = 8

You did it! You navigated the complexity and found the elegant truth. The final answer is $8$ . Keep this persistence; it is what defines a true JEE Advanced scholar.

Common Traps & FAQs

Wait, why does the sum of binomial coefficients equal

2^{n}

Think of the Binomial Theorem:

(1 + x)^{n} = \sum_{r = 0}^{n}^{n} C_{r} x^{r}

. If you set

x = 1

, the left side becomes

(1 + 1)^{n} = 2^{n}

, and the right side becomes the sum of all binomial coefficients. It is a beautiful, direct consequence of the theorem!

Why did we get

e - 2

for the second sum in series

a

Great catch! Remember that the standard expansion for

e

1 + \frac{1}{1 !} + \frac{1}{2 !} + \dots

. Our sum started at

n = 2

, which corresponds to

\frac{1}{2 !} + \frac{1}{3 !} + \dots

. Since the full series is

e

, we must subtract the first two terms (

1

and

\frac{1}{1 !}

) to get the remaining sum. Hence,

e - 1 - 1 = e - 2

Is there a faster way to handle the algebra in series

a

While it feels tedious, this 'algebraic surgery' is the most reliable method. By forcing the numerator to match the factors in the denominator's factorial, we turn a complex expression into a telescoping or standard series. It is a fundamental skill for JEE Advanced!

Enter Numerical Value:

Visualized Solution

Understanding Series $b$

Simplifying Series $b$

Evaluating Series $b$ using $e^{x}$

Analyzing Series $a$

Expanding the General Term $T_{n}$

Algebraic Manipulation of the Numerator

Splitting $T_{n}$ into Two Fractions

Simplifying the First Fraction

Simplifying the Second Fraction

Final Form of the Second Fraction

Combining to get Final $T_{n}$

Setting up the Sum for $a$

Evaluating the Infinite Sums

Calculating the Value of $a$

Final Calculation of $\frac{2 b}{a ^{2}}$

The Sigma Insight: Sum of Special Series

Analyzing the Setup

Decoding Series $b$

The Challenge of Series $a$

The Algebraic Surgery

The Grand Finale

Similar Questions

Let $\sum_{n = 0}^{\infty} \frac{n ^{3} (( 2 n )!) + ( 2 n - 1 ) ( n !)}{( n !) (( 2 n )!)} = a e + \frac{b}{e} + c$ , where $a, b, c \in Z$ and $e = \sum_{n = 0}^{\infty} \frac{1}{n !}$ . Then $a^{2} - b + c$ is equal to

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

If $0 < x < 1$ , then $\frac{3}{2} x^{2} + \frac{5}{3} x^{3} + \frac{7}{4} x^{4} + \dots$ , is equal to:

Let $a_{1} = b_{1} = 1$ and $a_{n} = a_{n - 1} + (n - 1), b_{n} = b_{n - 1} + a_{n - 1}, \forall n \geq 2$ . If $S = \sum_{n = 1}^{10} \frac{b _{n}}{2 ^{n}}$ and $T = \sum_{n = 1}^{8} \frac{n}{2 ^{n - 1}}$ , then $2^{7} (2 S - T)$ 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

Let $⟨ a_{n} ⟩$ be a sequence such that $a_{1} + a_{2} + \dots + a_{n} = \frac{n ^{2} + 3 n}{( n + 1 ) ( n + 2 )}$ . If $28 \sum_{k = 1}^{10} \frac{1}{a _{k}} = p_{1} p_{2} p_{3} \dots p_{m}$ , where $p_{1}, p_{2}, \dots, p_{m}$ are the first $m$ prime numbers, then $m$ is equal to

If $\frac{1}{1 ^{4}} + \frac{1}{2 ^{4}} + \frac{1}{3 ^{4}} + \dots \infty = \frac{π ^{4}}{90}$ , $\frac{1}{1 ^{4}} + \frac{1}{3 ^{4}} + \frac{1}{5 ^{4}} + \dots \infty = α$ , $\frac{1}{2 ^{4}} + \frac{1}{4 ^{4}} + \frac{1}{6 ^{4}} + \dots \infty = β$ , then $\frac{α}{β}$ is equal to

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 :

If $(\frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012}) - (\frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \frac{1}{6 \cdot 5} + \dots + \frac{1}{2024 \cdot 2023}) = \frac{1}{2024}$ , then $α$ is equal to

$(\frac{1}{3} + \frac{4}{7}) + (\frac{1}{3 ^{2}} + \frac{1}{3} \times \frac{4}{7} + \frac{4 ^{2}}{7 ^{2}}) + (\frac{1}{3 ^{3}} + \frac{1}{3 ^{2}} \times \frac{4}{7} + \frac{1}{3} \times \frac{4 ^{2}}{7 ^{2}} + \frac{4 ^{3}}{7 ^{3}}) + \dots$ upto infinite terms, is equal to

Enter Numerical Value:

Visualized Solution

Understanding Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Simplifying Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Evaluating Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ex

Analyzing Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Expanding the General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​

Algebraic Manipulation of the Numerator

Splitting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​ into Two Fractions

Simplifying the First Fraction

Simplifying the Second Fraction

Final Form of the Second Fraction

Combining to get Final .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​

Setting up the Sum for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Evaluating the Infinite Sums

Calculating the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a22b​

The Sigma Insight: Sum of Special Series

Analyzing the Setup

Decoding Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

The Challenge of Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

The Algebraic Surgery

The Grand Finale

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ∑n=0∞​(n!)((2n)!)n3((2n)!)+(2n−1)(n!)​=ae+eb​+c, where a,b,c∈Z and e=∑n=0∞​n!1​. Then a2−b+c is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of series 2!1​+4!1​+6!1​+… is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 0<x<1, then 23​x2+35​x3+47​x4+…, is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​=b1​=1 and an​=an−1​+(n−1),bn​=bn−1​+an−1​,∀n≥2. If S=∑n=110​2nbn​​ and T=∑n=18​2n−1n​, then 27(2S−T) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum 1+2!1+3​+3!1+3+5​+4!1+3+5+7​+… upto ∞ terms, is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 141​+241​+341​+…∞=90π4​, 141​+341​+541​+…∞=α, 241​+441​+641​+…∞=β, then βα​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Sn​=1⋅(n−1)+2⋅(n−2)+3⋅(n−3)+⋯+(n−1)⋅1,n≥4. The sum ∑n=4∞​(n!2Sn​​−(n−2)!1​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (α+11​+α+21​+⋯+α+10121​)−(2⋅11​+4⋅31​+6⋅51​+⋯+2024⋅20231​)=20241​, then α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (31​+74​)+(321​+31​×74​+7242​)+(331​+321​×74​+31​×7242​+7343​)+… upto infinite terms, is equal to

Understanding Series $b$

Simplifying Series $b$

Evaluating Series $b$ using $e^{x}$

Analyzing Series $a$

Expanding the General Term $T_{n}$

Splitting $T_{n}$ into Two Fractions

Combining to get Final $T_{n}$

Setting up the Sum for $a$

Calculating the Value of $a$

Final Calculation of $\frac{2 b}{a ^{2}}$

Decoding Series $b$

The Challenge of Series $a$

Let $\sum_{n = 0}^{\infty} \frac{n ^{3} (( 2 n )!) + ( 2 n - 1 ) ( n !)}{( n !) (( 2 n )!)} = a e + \frac{b}{e} + c$ , where $a, b, c \in Z$ and $e = \sum_{n = 0}^{\infty} \frac{1}{n !}$ . Then $a^{2} - b + c$ is equal to

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

If $0 < x < 1$ , then $\frac{3}{2} x^{2} + \frac{5}{3} x^{3} + \frac{7}{4} x^{4} + \dots$ , is equal to:

Let $a_{1} = b_{1} = 1$ and $a_{n} = a_{n - 1} + (n - 1), b_{n} = b_{n - 1} + a_{n - 1}, \forall n \geq 2$ . If $S = \sum_{n = 1}^{10} \frac{b _{n}}{2 ^{n}}$ and $T = \sum_{n = 1}^{8} \frac{n}{2 ^{n - 1}}$ , then $2^{7} (2 S - T)$ 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

If $\frac{1}{1 ^{4}} + \frac{1}{2 ^{4}} + \frac{1}{3 ^{4}} + \dots \infty = \frac{π ^{4}}{90}$ , $\frac{1}{1 ^{4}} + \frac{1}{3 ^{4}} + \frac{1}{5 ^{4}} + \dots \infty = α$ , $\frac{1}{2 ^{4}} + \frac{1}{4 ^{4}} + \frac{1}{6 ^{4}} + \dots \infty = β$ , then $\frac{α}{β}$ is equal to

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 :

If $(\frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012}) - (\frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \frac{1}{6 \cdot 5} + \dots + \frac{1}{2024 \cdot 2023}) = \frac{1}{2024}$ , then $α$ is equal to

$(\frac{1}{3} + \frac{4}{7}) + (\frac{1}{3 ^{2}} + \frac{1}{3} \times \frac{4}{7} + \frac{4 ^{2}}{7 ^{2}}) + (\frac{1}{3 ^{3}} + \frac{1}{3 ^{2}} \times \frac{4}{7} + \frac{1}{3} \times \frac{4 ^{2}}{7 ^{2}} + \frac{4 ^{3}}{7 ^{3}}) + \dots$ upto infinite terms, is equal to