JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: Suppose A and B are the coefficients of $3 0^{t h}$ and $1 2^{t h}$ terms respectively in the binomial expansion of $(1 + x)^{2 n - 1}$ . If $2 A = 5 B$ , then n is equal to :

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the expansion of

(1 + x)^{2 n - 1}

. The general term

T_{r + 1}

is given by

(r 2 n - 1) x^{r}

. Remember, the coefficient of the

k^{t h}

term corresponds to

r = k - 1

. Do not confuse the term number with the index

r

Hint 2

Hint 2 (The Arsenal): You are dealing with the ratio of two binomial coefficients. Instead of expanding the factorials into a massive polynomial, use the property

(r N) = \frac{N !}{r ! ( N - r )!}

. Focus on the ratio

\frac{A}{B} = \frac{5}{2}

and see how the

(2 n - 1)!

terms cancel out.

Hint 3

Hint 3 (The Execution): After substituting the factorial forms, you will arrive at

\frac{2}{5} = \frac{29 ! ( 2 n - 30 )!}{11 ! ( 2 n - 12 )!}

. Rearrange this to isolate the factorial ratios:

\frac{2}{5} = (\frac{29 !}{11 !}) \times (\frac{( 2 n - 30 )!}{( 2 n - 12 )!})

. Now, expand the larger factorials to reveal a sequence of consecutive integers that will simplify the equation significantly.

Visualized Solution

General Term of Binomial Expansion

Given expansion: $(1 + x)^{2 n - 1}$
General term formula: $T_{r + 1} =^{N} C_{r} x^{r}$
Here, the index is $N = 2 n - 1$
The coefficient of $T_{r + 1}$ is simply $^{2 n - 1} C_{r}$

Identifying Coefficients $A$ and $B$

For the $3 0^{t h}$ term, $r + 1 = 30 \Rightarrow r = 29$
Coefficient $A =^{2 n - 1} C_{29}$
For the $1 2^{t h}$ term, $r + 1 = 12 \Rightarrow r = 11$
Coefficient $B =^{2 n - 1} C_{11}$

Setting up the Equation $2 A = 5 B$

Given condition: $2 A = 5 B$
Substituting the values of $A$ and $B$ :
$2 \times^{2 n - 1} C_{29} = 5 \times^{2 n - 1} C_{11}$

Expanding Combinations with Factorials

Using the formula: $^{n} C_{r} = \frac{n !}{r ! ( n - r )!}$
Left Hand Side (LHS): $2 \times \frac{( 2 n - 1 )!}{29 ! ( 2 n - 1 - 29 )!} = 2 \times \frac{( 2 n - 1 )!}{29 ! ( 2 n - 30 )!}$
Right Hand Side (RHS): $5 \times \frac{( 2 n - 1 )!}{11 ! ( 2 n - 1 - 11 )!} = 5 \times \frac{( 2 n - 1 )!}{11 ! ( 2 n - 12 )!}$

Simplifying the Equation

Notice $(2 n - 1)!$ is common in the numerators on both sides.
Canceling $(2 n - 1)!$ gives:
$\frac{2}{29 ! ( 2 n - 30 )!} = \frac{5}{11 ! ( 2 n - 12 )!}$

Rearranging for Comparison

Let's group the constants on one side and factorials on the other.
Cross-multiplying to rearrange:
$\frac{2}{5} = \frac{29 ! ( 2 n - 30 )!}{11 ! ( 2 n - 12 )!}$
Separating the terms: $\frac{2}{5} = (\frac{29 !}{11 !}) \times (\frac{( 2 n - 30 )!}{( 2 n - 12 )!})$

Expanding Larger Factorials

Expanding $\frac{29 !}{11 !} = 29 \times 28 \times \dots \times 12$
Expanding $\frac{( 2 n - 12 )!}{( 2 n - 30 )!} = (2 n - 12) \times (2 n - 13) \times \dots \times (2 n - 29)$
Substituting back:
$\frac{2}{5} = \frac{29 \times 28 \times \dots \times 12}{( 2 n - 12 ) \times ( 2 n - 13 ) \times \dots \times ( 2 n - 29 )}$

Analyzing the Product Symmetry

Number of terms in numerator: $29 - 12 + 1 = 18$ terms.
Number of terms in denominator: $(2 n - 12) - (2 n - 29) + 1 = 18$ terms.
Both numerator and denominator have exactly $18$ consecutive decreasing integers.
Let's test if aligning the largest terms works: Assume $2 n - 12 = 30$ .

Testing the Hypothesis

If $2 n - 12 = 30$ , the denominator becomes: $30 \times 29 \times \dots \times 13$
The fraction becomes: $\frac{29 \times 28 \times \dots \times 12}{30 \times 29 \times \dots \times 13}$
Canceling common terms ( $29$ down to $13$ ):
Result $= \frac{12}{30} = \frac{2}{5}$
This perfectly matches the LHS!

Solving for $n$ and Conclusion

Since our assumption holds true, we set: $2 n - 12 = 30$
Transposing $12$ : $2 n = 30 + 12 \Rightarrow 2 n = 42$
Dividing by $2$ : $n = 21$
Final Answer: The value of $n$ is $21$ . (Option 2)

00:00 / 00:00

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The expansion of

(1 + x)^{2 n - 1}

follows the Binomial Theorem, where the general term is given by

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

To find the coefficients, we must remember that the

k^{t h}

term corresponds to

r = k - 1

. Therefore, for the

3 0^{t h}

term, we set

r = 29

, and for the

1 2^{t h}

term, we set

r = 11

The coefficients are defined as

A = (29 2 n - 1)

and

B = (11 2 n - 1)

. Always respect the index to avoid common pitfalls.

The Factorial Jungle

We are given the condition

2 A = 5 B

. Substituting our expressions for

A

and

B

, we obtain:

2 \times (29 2 n - 1) = 5 \times (11 2 n - 1)

Using the definition

(r N) = \frac{N !}{r ! ( N - r )!}

, we expand the equation:

2 \times \frac{( 2 n - 1 )!}{29 ! ( 2 n - 30 )!} = 5 \times \frac{( 2 n - 1 )!}{11 ! ( 2 n - 12 )!}

Since

(2 n - 1)!

appears on both sides, we can cancel it out. This simplifies the equation significantly:

\frac{2}{29 ! ( 2 n - 30 )!} = \frac{5}{11 ! ( 2 n - 12 )!}

The Symmetry Revelation

To solve for

n

, we rearrange the terms to group the constants and the variables:

\frac{2}{5} = \frac{29 !}{11 !} \times \frac{( 2 n - 30 )!}{( 2 n - 12 )!}

Note that

\frac{29 !}{11 !} = 29 \times 28 \times \dots \times 12

. Similarly, the ratio of the factorial terms involving

n

can be expressed as:

\frac{( 2 n - 12 )!}{( 2 n - 30 )!} = (2 n - 12) \times (2 n - 13) \times \dots \times (2 n - 29)

Both sides represent the product of 18 consecutive decreasing integers. By setting the largest term of the denominator,

2 n - 12

, equal to

30

, we satisfy the equality.

Final Calculation

Solving the linear equation:

2 n - 12 = 30

2 n = 42

$n = 21$

Common Traps & FAQs

Wait, why is the

3 0^{t h}

term

r = 29

and not

r = 30

? This always confuses me!

That is a classic trap! Always remember the general term formula:

T_{r + 1} = (r N) x^{r}

. The index

r

is always one less than the term number because the expansion starts with

T_{1}

(where

r = 0

). So, for the

3 0^{t h}

term,

r + 1 = 30

, which means

r = 29

. Keep this 'minus one' rule in your toolkit, and you will never fall for this again!

Why can't we just use the property

(r n) = (n - r n)

here?

You absolutely could! It is a valid property. However, in this specific problem, the upper index

N = 2 n - 1

is unknown. Using

(r N) = (N - r N)

would introduce

2 n - 1 - r

into your equation, which might make the algebra look more complex. The method of direct factorial expansion we used is more direct because it allows the

(2 n - 1)!

terms to cancel out immediately, leaving you with a much cleaner path to the solution.

Is there a faster way to solve the final equation without expanding all those terms?

Great intuition! You are looking for the 'JEE shortcut'. Once you reach the stage where you have a ratio of products of consecutive integers, you can often use symmetry or logical deduction. As we saw, both the numerator and denominator had 18 terms. By aligning the largest terms, we found that

2 n - 12

had to be

30

. Recognizing this symmetry is exactly how you save precious time in the exam hall!

Animated Solution for Mathematics - Binomial Theorem: Suppose A and B are the coefficients of $3 0^{t h}$ and $1 2^{t h}$ terms respectively in the binomial expansion of $(1 + x)^{2 n - 1}$ . If $2 A = 5 B$ , then n is equal to :

Select Answer:

Visualized Solution

General Term of Binomial Expansion

Identifying Coefficients $A$ and $B$

Setting up the Equation $2 A = 5 B$

Expanding Combinations with Factorials

Simplifying the Equation

Rearranging for Comparison

Expanding Larger Factorials

Analyzing the Product Symmetry

Testing the Hypothesis

Solving for $n$ and Conclusion

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Factorial Jungle

The Symmetry Revelation

Final Calculation

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Visualized Solution

General Term of Binomial Expansion

Identifying Coefficients .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Setting up the Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2A=5B

Expanding Combinations with Factorials

Simplifying the Equation

Rearranging for Comparison

Expanding Larger Factorials

Analyzing the Product Symmetry

Testing the Hypothesis

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

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Factorial Jungle

The Symmetry Revelation

Final Calculation

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