JEE Main 2021 (25 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: Let $n \in N$ and $[x]$ denote the greatest integer less than or equal to $x$ . If the sum of $(n + 1)$ terms $^{n} C_{0}, 3 \cdot^{n} C_{1}, 5 \cdot^{n} C_{2}, 7 \cdot^{n} C_{3}, \dots$ is equal to $2^{100} \cdot 101$ , then $2 [\frac{n - 1}{2}]$ is equal to ___

Enter Numerical Value:

Visualized Solution

Identify the General Term

Given series: $^{n} C_{0} + 3 \cdot^{n} C_{1} + 5 \cdot^{n} C_{2} + \dots$
The coefficients are $1, 3, 5, 7, \dots$ which form an Arithmetic Progression (A.P.).
General term of this A.P.: $1 + r (2) = 2 r + 1$
General term of the series: $T_{r} = (2 r + 1)^{n} C_{r}$

Express Sum using Summation

Let the sum of the series be $S$ .
Using summation notation from $r = 0$ to $n$ :
$S = \sum_{r = 0}^{n} (2 r + 1)^{n} C_{r}$

Split the Summation

Distribute the terms inside the summation:
$S = \sum_{r = 0}^{n} (2 r \cdot^{n} C_{r} + 1 \cdot^{n} C_{r})$
Split into two separate summations:
$S = 2 \sum_{r=0}^n r \cdot ^nC_r + \sum_{r=0}^n ^nC_r$

Apply Binomial Identities

Recall the standard binomial identity for the sum of coefficients: $\sum_{r=0}^n ^nC_r = 2^n$
Recall the identity involving $r$ : $\sum_{r = 0}^{n} r \cdot^{n} C_{r} = n \cdot 2^{n - 1}$
Substitute these identities back into our equation for $S$ .

Simplify the Sum Expression

Substitute the values: $S = 2 (n \cdot 2^{n - 1}) + 2^{n}$
Simplify the first term: $2 \cdot 2^{n - 1} = 2^{n}$
So, $S = n \cdot 2^{n} + 2^{n}$
Factor out $2^{n}$ : $S = (n + 1) 2^{n}$

Equate and Solve for $n$

We are given that $S = 101 \cdot 2^{100}$
Equating our result: $(n + 1) 2^{n} = 101 \cdot 2^{100}$
By direct comparison of the terms: $n = 100$

Setup the Final Expression

We need to find the value of $2 [\frac{n - 1}{2}]$
Substitute $n = 100$ into the expression.
$2 [\frac{100 - 1}{2}] = 2 [\frac{99}{2}]$

Evaluate Greatest Integer Function

Calculate the fraction: $\frac{99}{2} = 49.5$
Apply the greatest integer function: $[49.5] = 49$
Final Calculation: $2 \cdot 49 = 98$

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Decoding the Pattern

The first step in any great adventure is observation. Look at the coefficients:

1, 3, 5, 7, \dots

This is an Arithmetic Progression with a first term

a = 1

and a common difference

d = 2

. The general term of this progression is

2 r + 1

When we multiply this by the binomial coefficient

^{n} C_{r}

, we get the general term of our series:

T_{r} = (2 r + 1)^{n} C_{r}

By identifying this, we have already won half the battle. We have transformed a list of numbers into a precise mathematical expression.

The Art of Decomposition

Now, we write the sum

S = \sum_{r = 0}^{n} (2 r + 1)^{n} C_{r}

. We can distribute the

^{n} C_{r}

term to get

2 r \cdot^{n} C_{r} +^{n} C_{r}

By the linearity of summation, we can split this into two separate sums:

S = 2 \sum_{r=0}^n r \cdot ^nC_r + \sum_{r=0}^n ^nC_r

Suddenly, the monster is tamed. We are no longer looking at one complex series; we are looking at two standard, well-known identities.

The Binomial Toolkit

The sum of all binomial coefficients is:

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

The second identity, which is our secret weapon, is:

r = 0 \sum n r \cdot^{n} C_{r} = n \cdot 2^{n - 1}

By substituting these into our split equation, we get:

S = 2 (n \cdot 2^{n - 1}) + 2^{n}

Look at the elegance of the simplification! Since

2 \cdot 2^{n - 1} = 2^{n}

, the expression becomes:

S = n \cdot 2^{n} + 2^{n} = (n + 1) 2^{n}

The Final Reveal

We are given that

S = 101 \cdot 2^{100}

. By equating our derived result

(n + 1) 2^{n} = 101 \cdot 2^{100}

, we find that

n = 100

The question asks for the value of

2 [\frac{n - 1}{2}]

. Substituting

n = 100

2 [\frac{100 - 1}{2}] = 2 [\frac{99}{2}] = 2 [49.5]

The greatest integer of

49.5

49

. Multiplying by

2

gives us the final result: 98

Common Traps & FAQs

Wait, why do we split the summation into two parts? Can't we just solve it as one big block?

That is a great question! While you *could* try to solve it as one block, it becomes algebraically messy. By splitting the summation, we isolate the variable

r

from the constant

1

. This allows us to apply standard, well-known binomial identities to each part individually, turning a complex problem into two simple, manageable pieces. It is all about reducing cognitive load!

How do I derive

\sum_{r = 0}^{n} r \cdot^{n} C_{r} = n \cdot 2^{n - 1}

if I forget it during the exam?

Panic not! This is a classic JEE trick. Start with the binomial expansion

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

. Differentiate both sides with respect to

x

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

. Now, simply set

x = 1

. The left side becomes

n (2)^{n - 1}

, and the right side becomes

\sum r \cdot^{n} C_{r}

. Voila! You have derived the identity from first principles.

What exactly does the notation

[\frac{n - 1}{2}]

mean? Is it just a bracket?

That is the Greatest Integer Function, often called the 'floor' function. It means you take the value inside the brackets and round it down to the nearest integer. For example, if you have

[49.5]

, the greatest integer less than or equal to

49.5

49

. It is a common way in JEE problems to test if you can handle the final step with precision!

Enter Numerical Value:

Visualized Solution

Identify the General Term

Express Sum using Summation

Split the Summation

Apply Binomial Identities

Simplify the Sum Expression

Equate and Solve for $n$

Setup the Final Expression

Evaluate Greatest Integer Function

The Sigma Insight: Properties of Binomial Coefficients

Decoding the Pattern

The Art of Decomposition

The Binomial Toolkit

The Final Reveal

Similar Questions

Let $[x]$ denote greatest integer less than or equal to $x$ . If for $n \in N$ , $(1 - x + x^{3})^{n} = \sum_{j = 0}^{3 n} a_{j} x^{j}$ , then $\sum_{j = 0}^{[\frac{3 n}{2}]} a_{2 j} + 4 \sum_{j = 0}^{[\frac{3 n - 1}{2}]} a_{2 j + 1}$ is equal to :

Let $S_{n} = 1 + q + q^{2} + \dots + q^{n}$ and $T_{n} = 1 + (\frac{q + 1}{2}) + (\frac{q + 1}{2})^{2} + \dots + (\frac{q + 1}{2})^{n}$ where $q$ is a real number and $q \neq = 1$ . If $^{101} C_{1} +^{101} C_{2} \cdot S_{1} + \dots +^{101} C_{101} \cdot S_{100} = α T_{100}$ then $α$ is equal to:

The value of $\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots + \frac{^{100} C _{100}}{101}$ is :

If $1 + (^{49} C_{0} +^{49} C_{1} + \dots +^{49} C_{49} + 1) (^{50} C_{2} +^{50} C_{4} + \dots +^{50} C_{50})$ is equal to $2^{n} \cdot m$ , where $m$ is odd, then $n + m$ is equal to ______.

If $n \geq 2$ is a positive integer, then the sum of the series $^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + ... +^{n} C_{2})$ is :

Find the sum of the series : $\sum_{r = 0}^{n} (- 1)^{r}^{n} C_{r} [\frac{1}{2 ^{r}} + \frac{3 ^{r}}{2 ^{2 r}} + \frac{7 ^{r}}{2 ^{3 r}} + \frac{1 5 ^{r}}{2 ^{4 r}} \dots up to m terms]$ .

If $\frac{1}{n + 1}^{n} C_{n} + \frac{1}{n}^{n} C_{n - 1} + \dots + \frac{1}{2}^{n} C_{1} +^{n} C_{0} = \frac{1023}{10}$ then $n$ is equal to

Let $S = \frac{1}{25 !} + \frac{1}{3 ! 23 !} + \frac{1}{5 ! 21 !} + \dots$ up to 13 terms. If $13 S = \frac{2 ^{k}}{n !}, k \in N$ , then $n + k$ is equal to

The sum, of the coefficients of the first 50 terms in the binomial expansion of $(1 - x)^{100}$ , is equal to

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ ; $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}, q \neq = 1$ Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

Enter Numerical Value:

Visualized Solution

Identify the General Term

Express Sum using Summation

Split the Summation

Apply Binomial Identities

Simplify the Sum Expression

Equate and Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Setup the Final Expression

Evaluate Greatest Integer Function

The Sigma Insight: Properties of Binomial Coefficients

Decoding the Pattern

The Art of Decomposition

The Binomial Toolkit

The Final Reveal

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of 51100C50​​+52100C51​​+⋯+101100C100​​ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1+(49C0​+49C1​+⋯+49C49​+1)(50C2​+50C4​+⋯+50C50​) is equal to 2n⋅m, where m is odd, then n+m is equal to ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If n≥2 is a positive integer, then the sum of the series n+1C2​+2(2C2​+3C2​+4C2​+...+nC2​) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the sum of the series : ∑r=0n​(−1)rnCr​[2r1​+22r3r​+23r7r​+24r15r​… up to m terms].

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If n+11​nCn​+n1​nCn−1​+⋯+21​nC1​+nC0​=101023​ then n is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S=25!1​+3!23!1​+5!21!1​+… up to 13 terms. If 13S=n!2k​,k∈N, then n+k is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum, of the coefficients of the first 50 terms in the binomial expansion of (1−x)100, is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given sn​=1+q+q2+....+qn; Sn​=1+2q+1​+(2q+1​)2+....+(2q+1​)n,q=1 Prove that n+1C1​+n+1C2​s1​+n+1C3​s2​+....+n+1Cn+1​sn​=2nSn​

Equate and Solve for $n$

The value of $\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots + \frac{^{100} C _{100}}{101}$ is :

If $1 + (^{49} C_{0} +^{49} C_{1} + \dots +^{49} C_{49} + 1) (^{50} C_{2} +^{50} C_{4} + \dots +^{50} C_{50})$ is equal to $2^{n} \cdot m$ , where $m$ is odd, then $n + m$ is equal to ______.

If $n \geq 2$ is a positive integer, then the sum of the series $^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + ... +^{n} C_{2})$ is :

Find the sum of the series : $\sum_{r = 0}^{n} (- 1)^{r}^{n} C_{r} [\frac{1}{2 ^{r}} + \frac{3 ^{r}}{2 ^{2 r}} + \frac{7 ^{r}}{2 ^{3 r}} + \frac{1 5 ^{r}}{2 ^{4 r}} \dots up to m terms]$ .

If $\frac{1}{n + 1}^{n} C_{n} + \frac{1}{n}^{n} C_{n - 1} + \dots + \frac{1}{2}^{n} C_{1} +^{n} C_{0} = \frac{1023}{10}$ then $n$ is equal to

Let $S = \frac{1}{25 !} + \frac{1}{3 ! 23 !} + \frac{1}{5 ! 21 !} + \dots$ up to 13 terms. If $13 S = \frac{2 ^{k}}{n !}, k \in N$ , then $n + k$ is equal to

The sum, of the coefficients of the first 50 terms in the binomial expansion of $(1 - x)^{100}$ , is equal to

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ ; $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}, q \neq = 1$ Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$