JEE Main 2020 - 2 Sep (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: For a positive integer $n$ , $(1 + \frac{1}{x})^{n}$ is expanded in increasing powers of $x$ . If three consecutive coefficients in this expansion are in the ratio, $2 : 5 : 12$ , then $n$ is equal to

Q: Why can't we just expand the whole thing? It seems like a lot of work.

Expanding $(1+\frac{1}{x})^n$ for an unknown $n$ is impossible because we don't know the degree! The beauty of the Binomial Theorem lies in the properties of its coefficients. By using the ratio property $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}$, we bypass the need for expansion entirely and jump straight to the algebraic relationship between $n$ and $r$.

Q: Is $r$ always an integer? What if I get a fraction?

In the context of binomial coefficients $\binom{n}{r}$, $r$ must be a non-negative integer because it represents the position of a term in the expansion. If your calculation yields a non-integer for $r$, it is a clear signal to re-check your algebraic manipulation of the ratio equations. In this problem, $r$ must come out to be exactly $34$.

Enter Numerical Value:

Visualized Solution

Introduction to the Problem

Given expression: $(1 + \frac{1}{x})^{n}$
Ratio of three consecutive coefficients: $2 : 5 : 12$
Objective: Find the value of $n$ .

Analyzing the Expansion

Expanding $(1 + \frac{1}{x})^{n}$ in increasing powers of $x$ :
$(1 + \frac{1}{x})^{n} = (0 n) x^{- n} + (1 n) x^{- n + 1} + \dots + (n n) x^{0}$
The sequence of coefficients is $(0 n), (1 n), \dots, (n n)$ .

Defining Consecutive Coefficients

Let the three consecutive coefficients be $(r - 1 n), (r n), and (r + 1 n)$ .
Given ratio: $(r - 1 n) : (r n) : (r + 1 n) = 2 : 5 : 12$

Setting up the First Ratio

First part of the ratio: $\frac{( r - 1 n )}{( r n )} = \frac{2}{5}$
Taking reciprocal for convenience: $\frac{( r n )}{( r - 1 n )} = \frac{5}{2}$

Applying the Ratio Formula

Using the standard property: $\frac{( r n )}{( r - 1 n )} = \frac{n - r + 1}{r}$
Equating both expressions: $\frac{n - r + 1}{r} = \frac{5}{2}$

Deriving the First Equation

Cross-multiplying: $2 (n - r + 1) = 5 r$
Expanding: $2 n - 2 r + 2 = 5 r$
Rearranging terms: $2 n - 7 r = - 2$ (Equation 1)

Setting up the Second Ratio

Second part of the ratio: $\frac{( r n )}{( r + 1 n )} = \frac{5}{12}$
Taking reciprocal: $\frac{( r + 1 n )}{( r n )} = \frac{12}{5}$

Applying the Formula Again

Using the property: $\frac{( r + 1 n )}{( r n )} = \frac{n - ( r + 1 ) + 1}{r + 1} = \frac{n - r}{r + 1}$
Equating both: $\frac{n - r}{r + 1} = \frac{12}{5}$

Deriving the Second Equation

Cross-multiplying: $5 (n - r) = 12 (r + 1)$
Expanding: $5 n - 5 r = 12 r + 12$
Rearranging terms: $5 n - 17 r = 12$ (Equation 2)

The System of Equations

We now have a system of two linear equations:
1) $2 n - 7 r = - 2$
2) $5 n - 17 r = 12$

Solving for r

Multiply Eq (1) by $5$ : $10 n - 35 r = - 10$
Multiply Eq (2) by $2$ : $10 n - 34 r = 24$
Subtracting the equations: $(10 n - 34 r) - (10 n - 35 r) = 24 - (- 10)$
Result: $r = 34$

Finding the Final Value of n

Substitute $r = 34$ into $2 n - 7 r = - 2$ :
$2 n - 7 (34) = - 2$
$2 n - 238 = - 2$
$2 n = 236 ⟹ n = 118$

Conclusion and Summary

Final value of $n$ is $118$ .
Key Takeaway: The ratio of consecutive coefficients $\frac{( r n )}{( r - 1 n )} = \frac{n - r + 1}{r}$ is a fundamental tool.
Always convert ratio problems into a system of linear equations.

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The Sigma Insight: Properties of Binomial Coefficients

Solution Diagram

Analyzing the Setup

The heart of this problem lies in the coefficients of the binomial expansion. We are given three consecutive coefficients in the ratio

2 : 5 : 12

Let these coefficients be represented by the combinations

(r - 1 n)

(r n)

, and

(r + 1 n)

. The problem states:

(r - 1 n) : (r n) : (r + 1 n) = 2 : 5 : 12

Decoding the Ratio

To solve this, we break the triple ratio into two manageable equations. We start with the first two terms:

\frac{( r - 1 n )}{( r n )} = \frac{2}{5} ⟹ \frac{( r n )}{( r - 1 n )} = \frac{5}{2}

We invoke the fundamental binomial ratio property:

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

Substituting this into our equation, we obtain:

\frac{n - r + 1}{r} = \frac{5}{2}

Cross-multiplying yields

2 (n - r + 1) = 5 r

, which simplifies to our first linear equation:

2 n - 7 r = - 2

The Second Half of the Journey

Now, we address the second part of the ratio:

\frac{( r n )}{( r + 1 n )} = \frac{5}{12} ⟹ \frac{( r + 1 n )}{( r n )} = \frac{12}{5}

Applying the ratio property again, specifically for the index

r + 1

, we get:

\frac{n - ( r + 1 ) + 1}{r + 1} = \frac{n - r}{r + 1} = \frac{12}{5}

Cross-multiplying yields

5 (n - r) = 12 (r + 1)

, which simplifies to:

5 n - 17 r = 12

The Final Resolution

We now have a system of two linear equations:

2 n - 7 r = - 2

5 n - 17 r = 12

To eliminate

n

, we multiply the first equation by

5

and the second by

2

10 n - 35 r = - 10

10 n - 34 r = 24

Subtracting the first equation from the second, the

10 n

terms vanish, leaving us with:

r = 34

Substituting

r = 34

back into the first equation:

2 n - 7 (34) = - 2

2 n - 238 = - 2

2 n = 236

Thus, the final values are $n = 118$ and $r = 34$ .

Common Traps & FAQs

Wait, why do we use

(r - 1 n) : (r n) : (r + 1 n)

? How do we know the order?

Great observation! The problem states 'three consecutive coefficients' are in the ratio

2 : 5 : 12

. In any binomial expansion, the coefficients increase and then decrease. Since

2 < 5 < 12

is not possible for a single sequence, we must interpret the ratio as the relative values of these three consecutive terms. The order is fixed by the problem statement itself, and we simply assign the indices

r - 1, r, r + 1

to represent their consecutive nature.

Why can't we just expand the whole thing? It seems like a lot of work.

Expanding

(1 + \frac{1}{x})^{n}

for an unknown

n

is impossible because we don't know the degree! The beauty of the Binomial Theorem lies in the properties of its coefficients. By using the ratio property

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

, we bypass the need for expansion entirely and jump straight to the algebraic relationship between

n

and

r

r

always an integer? What if I get a fraction?

In the context of binomial coefficients

(r n)

r

must be a non-negative integer because it represents the position of a term in the expansion. If your calculation yields a non-integer for

r

, it is a clear signal to re-check your algebraic manipulation of the ratio equations. In this problem,

r

must come out to be exactly

34

Animated Solution for Mathematics - Binomial Theorem: For a positive integer $n$ , $(1 + \frac{1}{x})^{n}$ is expanded in increasing powers of $x$ . If three consecutive coefficients in this expansion are in the ratio, $2 : 5 : 12$ , then $n$ is equal to

Enter Numerical Value:

Visualized Solution

Introduction to the Problem

Analyzing the Expansion

Defining Consecutive Coefficients

Setting up the First Ratio

Applying the Ratio Formula

Deriving the First Equation

Setting up the Second Ratio

Applying the Formula Again

Deriving the Second Equation

The System of Equations

Solving for r

Finding the Final Value of n

Conclusion and Summary

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Decoding the Ratio

The Second Half of the Journey

The Final Resolution

Similar Questions

The coefficients of three consecutive terms of $(1 + x)^{n + 5}$ are in the ratio $5 : 10 : 14$ . Then $n = .........$

The sum of the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{n + 2}$ , which are in the ratio $1 : 3 : 5$ , is equal to

$r$ and $n$ are positive integers $r > 1, n > 2$ and coefficient of $(r + 2)^{t h}$ term and $3 r^{t h}$ term in the expansion of $(1 + x)^{2 n}$ are equal, then $n$ equals

If for some positive integer $n$ , the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{n + 5}$ are in the ratio $5 : 10 : 14$ , then the largest coefficient in this expansion is :

If the coefficients of $x^{4}, x^{5}$ and $x^{6}$ in the expansion of $(1 + x)^{n}$ are in the arithmetic progression, then the maximum value of $n$ is:

Let the coefficients of three consecutive terms in the binomial expansion of $(1 + 2 x)^{n}$ be in the ratio $2 : 5 : 8$ . Then the coefficient of the term, which is in the middle of these three terms, is ______.

Let $n$ be positive integer. If the coefficients of 2nd, 3rd, and 4th terms in the expansion of $(1 + x)^{n}$ are in A.P., then the value of $n$ is .........

Let $m$ be the smallest positive integer such that the coefficient of $x^{2}$ in the expansion of $(1 + x)^{2} + (1 + x)^{3} + ... + (1 + x)^{49} + (1 + m x)^{50}$ is $(3 n + 1)^{51} C_{3}$ for some positive integer $n$ . Then the value of $n$ is .........

If some three consecutive in the binomial expansion of $(x + 1)^{n}$ in powers of $x$ are in the ratio $2 : 15 : 70$ , then the average of these three coefficients is :-

If the sum of the coefficients of all even powers of $x$ in the product $(1 + x + x^{2} + \dots + x^{2 n}) (1 - x + x^{2} - \dots + x^{2 n})$ is 61, then $n$ is equal to

Enter Numerical Value:

Visualized Solution

Introduction to the Problem

Analyzing the Expansion

Defining Consecutive Coefficients

Setting up the First Ratio

Applying the Ratio Formula

Deriving the First Equation

Setting up the Second Ratio

Applying the Formula Again

Deriving the Second Equation

The System of Equations

Solving for r

Finding the Final Value of n

Conclusion and Summary

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Decoding the Ratio

The Second Half of the Journey

The Final Resolution

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let n be positive integer. If the coefficients of 2nd, 3rd, and 4th terms in the expansion of (1+x)n are in A.P., then the value of n is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If some three consecutive in the binomial expansion of (x+1)n in powers of x are in the ratio 2:15:70, then the average of these three coefficients is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the coefficients of all even powers of x in the product (1+x+x2+⋯+x2n)(1−x+x2−⋯+x2n) is 61, then n is equal to

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$r$ and $n$ are positive integers $r > 1, n > 2$ and coefficient of $(r + 2)^{t h}$ term and $3 r^{t h}$ term in the expansion of $(1 + x)^{2 n}$ are equal, then $n$ equals

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