JEE Main 2022 (29 July Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $(42 + \frac{1}{4 3})^{n}$ , in the increasing powers of $\frac{1}{4 3}$ be $46 : 1$ . If the sixth term from the beginning is $\frac{α}{4 3}$ , then $α$ is equal to ______.

Q: Wait, why is the $k^{th}$ term from the end the $(n-k+2)^{th}$ term from the beginning? This feels counter-intuitive.

Think about a small example. If $n=4$, we have 5 terms ($T_1, T_2, T_3, T_4, T_5$). The 2nd term from the end is $T_4$. Using the formula: $n-k+2 = 4-2+2 = 4$. It works! The $+2$ accounts for the fact that we are counting the term itself and the offset from the total count $n+1$. It is a standard mapping technique that saves you from counting backwards manually.

Q: Why do we use $r=n-4$ for the term from the end?

In the general term formula $T_{r+1}$, the index $r$ is always one less than the term number. If we are looking for the $(n-3)^{th}$ term, we must set $r+1 = n-3$, which implies $r = n-4$. Always trust the formula over your intuition when dealing with indices!

Enter Numerical Value:

Progressive Hints

Hint 1

The symmetry of binomial coefficients is your best friend. Remember that

^{n} C_{r} =^{n} C_{n - r}

. Visualize the expansion: the

k^{t h}

term from the end is equivalent to the

(n - k + 2)^{t h}

term from the beginning. This simple shift in perspective turns a daunting problem into a manageable one.

Hint 2

Your primary tool is the general term formula:

T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}

. When dealing with fractional powers like

42

and

\frac{1}{4 3}

, do not panic. Treat them as exponents:

2^{\frac{1}{4}}

and

3^{- \frac{1}{4}}

. The laws of indices are your allies here.

Hint 3

Start by setting up the ratio equation:

\frac{T _{5}}{T _{n - 3}} = 6^{\frac{1}{4}}

. For

T_{5}

, use

r = 4

. For

T_{n - 3}

, use

r = n - 4

. Once you write this out, you will see the binomial coefficients

^{n} C_{4}

and

^{n} C_{n - 4}

cancel out beautifully, leaving you with a clean equation in terms of

n

Visualized Solution

The Binomial Expansion Sequence

Given expansion: $(42 + \frac{1}{4 3})^{n}$
Let $a = 2^{\frac{1}{4}}$ and $b = 3^{- \frac{1}{4}}$
The total number of terms in this expansion is $n + 1$ .

General Term Formula

The general term $T_{r + 1}$ in the expansion of $(a + b)^{n}$ is given by:
$T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$

Fifth Term from Beginning ( $T_{5}$ )

For the 5th term from the beginning, we set $r = 4$ .
$T_{5} =^{n} C_{4} (2^{\frac{1}{4}})^{n - 4} (3^{- \frac{1}{4}})^{4}$
Simplifying the power of 3: $(3^{- \frac{1}{4}})^{4} = 3^{- 1}$

Concept: Term from the End

The $k^{t h}$ term from the end in $(a + b)^{n}$ is the $(n - k + 2)^{t h}$ term from the beginning.
For $k = 5$ , the term is $T_{n - 5 + 2} = T_{n - 3}$

Fifth Term from End ( $T_{n - 3}$ )

For $T_{n - 3}$ , we set $r = n - 4$ .
$T_{n - 3} =^{n} C_{n - 4} (2^{\frac{1}{4}})^{n - (n - 4)} (3^{- \frac{1}{4}})^{n - 4}$
$T_{n - 3} =^{n} C_{n - 4} (2^{\frac{1}{4}})^{4} (3^{- \frac{1}{4}})^{n - 4}$

Setting up the Ratio

Given Ratio: $\frac{T _{5}}{T _{n - 3}} = 46 = 6^{\frac{1}{4}}$
Recall the property of combinations: $^{n} C_{r} =^{n} C_{n - r}$
Therefore, $^{n} C_{4} =^{n} C_{n - 4}$

Simplifying the Ratio

After canceling the binomial coefficients, we divide the terms:
$\frac{( 2 ^{\frac{1}{4}} ) ^{n - 4} \cdot 3 ^{- 1}}{2 ^{1} \cdot ( 3 ^{- \frac{1}{4}} ) ^{n - 4}} = 6^{\frac{1}{4}}$
Using laws of indices $\frac{x ^{m}}{x ^{n}} = x^{m - n}$

Combining Powers of 2 and 3

For base 2: $\frac{2 ^{\frac{n - 4}{4}}}{2 ^{1}} = 2^{\frac{n - 4}{4} - 1} = 2^{\frac{n - 8}{4}}$
For base 3: $\frac{3 ^{- 1}}{3 ^{- \frac{n - 4}{4}}} = 3^{- 1 + \frac{n - 4}{4}} = 3^{\frac{n - 8}{4}}$
Combined: $2^{\frac{n - 8}{4}} \cdot 3^{\frac{n - 8}{4}} = (2 \cdot 3)^{\frac{n - 8}{4}} = 6^{\frac{n - 8}{4}}$

Equating Exponents to Find $n$

We have: $6^{\frac{n - 8}{4}} = 6^{\frac{1}{4}}$
Equating the exponents: $\frac{n - 8}{4} = \frac{1}{4}$
$n - 8 = 1 ⟹ n = 9$

Setting up the Sixth Term ( $T_{6}$ )

The question asks for the 6th term from the beginning, $T_{6}$ .
For $T_{6}$ , we use $n = 9$ and $r = 5$ .
$T_{6} =^{9} C_{5} (2^{\frac{1}{4}})^{9 - 5} (3^{- \frac{1}{4}})^{5}$

Evaluating the Sixth Term

Calculate $^{9} C_{5} = \frac{9 !}{5 ! 4 !} = 126$
Simplify powers: $(2^{\frac{1}{4}})^{4} = 2^{1} = 2$
$(3^{- \frac{1}{4}})^{5} = 3^{- \frac{5}{4}}$
$T_{6} = 126 \cdot 2 \cdot 3^{- \frac{5}{4}} = 252 \cdot 3^{- \frac{5}{4}}$

Final Evaluation of $α$

Rewrite $T_{6}$ : $252 \cdot 3^{- \frac{5}{4}} = \frac{252}{3 ^{\frac{5}{4}}}$
Split the denominator: $3^{\frac{5}{4}} = 3^{1} \cdot 3^{\frac{1}{4}} = 3 \cdot 43$
$T_{6} = \frac{252}{3 \cdot 4 3} = \frac{84}{4 3}$
Comparing with $\frac{α}{4 3}$ , we get $α = 84$ .

00:00 / 00:00

The Sigma Insight: General Term and Middle Term

Solution Diagram

Analyzing the Setup

The expansion is given by

(42 + \frac{1}{4 3})^{n}

. We are tasked with finding the value of

n

and subsequently the sixth term of the expansion, given the ratio of the fifth term from the beginning to the fifth term from the end.

The total number of terms in the expansion is

n + 1

. We must utilize the property of binomial symmetry to relate terms from the beginning and the end.

The Symmetry of Terms

The

k^{t h}

term from the end is equivalent to the

(n - k + 2)^{t h}

term from the beginning. For the fifth term from the end (

k = 5

), this corresponds to the

(n - 5 + 2)^{t h} = (n - 3)^{t h}

term from the beginning.

Using the general term formula

T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}

, we identify the two terms: The fifth term from the beginning ( $T_{5}$ ) occurs at $r = 4$ :
$T_{5} =^{n} C_{4} (2^{1/4})^{n - 4} (3^{- 1/4})^{4}$
The fifth term from the end (

T_{n - 3}

) occurs at

r = n - 4

T_{n - 3} =^{n} C_{n - 4} (2^{1/4})^{4} (3^{- 1/4})^{n - 4}

The Master Equation

We are given the ratio

\frac{T _{5}}{T _{n - 3}} = 66

. Since

^{n} C_{4} =^{n} C_{n - 4}

, the binomial coefficients cancel out entirely.

The ratio simplifies to:

\frac{( 2 ^{1/4} ) ^{n - 4} ( 3 ^{- 1/4} ) ^{4}}{( 2 ^{1/4} ) ^{4} ( 3 ^{- 1/4} ) ^{n - 4}} = 6^{1/4}

Applying the laws of exponents, we group the bases:

\frac{2 ^{(n - 4) /4}}{2 ^{1}} \cdot \frac{3 ^{- 1}}{3 ^{- (n - 4) /4}} = 6^{1/4}

2^{(n - 8) /4} \cdot 3^{(n - 8) /4} = 6^{1/4}

This simplifies to the elegant form:

6^{(n - 8) /4} = 6^{1/4}

Solving for $n$

Equating the exponents, we have:

\frac{n - 8}{4} = \frac{1}{4}

\Rightarrow n - 8 = 1

\Rightarrow * * n = 9 * *

Final Calculation

With

n = 9

, we now calculate the sixth term (

T_{6}

), which corresponds to

r = 5

T_{6} =^{9} C_{5} (2^{1/4})^{9 - 5} (3^{- 1/4})^{5}

T_{6} = 126 \cdot (2^{1/4})^{4} \cdot (3^{- 1/4})^{5}

Simplifying the powers:

T_{6} = 126 \cdot 2^{1} \cdot 3^{- 5/4}

T_{6} = 252 \cdot \frac{1}{3 \cdot 3 ^{1/4}}

T_{6} = \frac{84}{4 3}

Comparing this to the form

α \cdot \frac{1}{4 3}

, we find that $α = 84$ .

Common Traps & FAQs

Wait, why is the

k^{t h}

term from the end the

(n - k + 2)^{t h}

term from the beginning? This feels counter-intuitive.

Think about a small example. If

n = 4

, we have 5 terms (

T_{1}, T_{2}, T_{3}, T_{4}, T_{5}

). The 2nd term from the end is

T_{4}

. Using the formula:

n - k + 2 = 4 - 2 + 2 = 4

. It works! The

+ 2

accounts for the fact that we are counting the term itself and the offset from the total count

n + 1

. It is a standard mapping technique that saves you from counting backwards manually.

Why do we use

r = n - 4

for the term from the end?

In the general term formula

T_{r + 1}

, the index

r

is always one less than the term number. If we are looking for the

(n - 3)^{t h}

term, we must set

r + 1 = n - 3

, which implies

r = n - 4

. Always trust the formula over your intuition when dealing with indices!

How do I handle the fractional powers without getting lost in the algebra?

The secret is to keep the bases separate. Don't try to multiply

2

and

3

until the very end. Keep the powers of

2

together and the powers of

3

together. When you divide, subtract the exponents. You will find that the exponents for both bases eventually become identical, which is the 'Aha!' moment of the problem.

Enter Numerical Value:

Visualized Solution

The Binomial Expansion Sequence

General Term Formula

Fifth Term from Beginning ( $T_{5}$ )

Concept: Term from the End

Fifth Term from End ( $T_{n - 3}$ )

Setting up the Ratio

Simplifying the Ratio

Combining Powers of 2 and 3

Equating Exponents to Find $n$

Setting up the Sixth Term ( $T_{6}$ )

Evaluating the Sixth Term

Final Evaluation of $α$

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Symmetry of Terms

The Master Equation

Solving for $n$

Final Calculation

Similar Questions

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(42 + \frac{1}{4 3})^{n}$ is $6 : 1$ , then the third term from the beginning is:

A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of $(2^{1/3} + \frac{1}{2 ( 3 ) ^{1/3}})^{10}$ is :

If the constant term in the binomial expansion of $(\frac{x ^{5/2}}{2} - \frac{4}{x ^{l}})^{9}$ is $- 84$ and the Coefficient of $x^{- 3 l}$ is $2^{α} β$ , where $β < 0$ is an odd number, Then $∣ α l - β ∣$ is equal to ______.

In the expansion of $(32 + \frac{1}{3 3})^{n}, n \in N$ , if the ratio of $1 5^{t h}$ term from the beginning to the $1 5^{t h}$ term from the end is $\frac{1}{6}$ , then the value of $^{n} C_{3}$ is:

If the sum of the coefficients of all the positive even powers of $x$ in the binomial expansion of $(2 x^{3} + \frac{3}{x})^{10}$ is $5^{10} - β \cdot 3^{9}$ , then $β$ is equal to ____.

The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1 + α x)^{4}$ and of $(1 - α x)^{6}$ is the same if $α$ equals

In the binomial expansion of $(a - b)^{n}, n \geq 5$ , the sum of $5^{t h}$ and $6^{t h}$ terms is zero, then $a / b$ equals

The Coefficient of $x^{- 6}$ in the expansion of $(\frac{4 x}{5} + \frac{5}{2 x ^{2}})^{9}$ is ______.

In the binomial expansion of $(a - b)^{n}, n \geq 5$ , the sum of the $5^{t h}$ and $6^{t h}$ terms is zero. Then $a / b$ equals

Let $α$ be the constant term in the binomial expansion of $(x - \frac{6}{x ^{\frac{3}{2}}})^{n}, n \leq 15$ . If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of $x^{- n}$ is $λ α$ , then $λ$ is equal to ————————.

Enter Numerical Value:

Visualized Solution

The Binomial Expansion Sequence

General Term Formula

Fifth Term from Beginning ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T5​)

Concept: Term from the End

Fifth Term from End ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn−3​)

Setting up the Ratio

Simplifying the Ratio

Combining Powers of 2 and 3

Equating Exponents to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Setting up the Sixth Term ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T6​)

Evaluating the Sixth Term

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

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Symmetry of Terms

The Master Equation

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

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (42​+43​1​)n is 6​:1, then the third term from the beginning is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of (21/3+2(3)1/31​)10 is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the expansion of (32​+33​1​)n,n∈N, if the ratio of 15th term from the beginning to the 15th term from the end is 61​, then the value of nC3​ 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 the positive even powers of x in the binomial expansion of (2x3+x3​)10 is 510−β⋅39, then β is equal to ____.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of the middle term in the binomial expansion in powers of x of (1+αx)4 and of (1−αx)6 is the same if α equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the binomial expansion of (a−b)n,n≥5, the sum of 5th and 6th terms is zero, then a/b equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Coefficient of x−6 in the expansion of (54x​+2x25​)9 is ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the binomial expansion of (a−b)n,n≥5, the sum of the 5th and 6th terms is zero. Then a/b equals

Fifth Term from Beginning ( $T_{5}$ )

Fifth Term from End ( $T_{n - 3}$ )

Equating Exponents to Find $n$

Setting up the Sixth Term ( $T_{6}$ )

Final Evaluation of $α$

Solving for $n$

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(42 + \frac{1}{4 3})^{n}$ is $6 : 1$ , then the third term from the beginning is:

A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of $(2^{1/3} + \frac{1}{2 ( 3 ) ^{1/3}})^{10}$ is :

In the expansion of $(32 + \frac{1}{3 3})^{n}, n \in N$ , if the ratio of $1 5^{t h}$ term from the beginning to the $1 5^{t h}$ term from the end is $\frac{1}{6}$ , then the value of $^{n} C_{3}$ is:

If the sum of the coefficients of all the positive even powers of $x$ in the binomial expansion of $(2 x^{3} + \frac{3}{x})^{10}$ is $5^{10} - β \cdot 3^{9}$ , then $β$ is equal to ____.

The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1 + α x)^{4}$ and of $(1 - α x)^{6}$ is the same if $α$ equals

In the binomial expansion of $(a - b)^{n}, n \geq 5$ , the sum of $5^{t h}$ and $6^{t h}$ terms is zero, then $a / b$ equals

The Coefficient of $x^{- 6}$ in the expansion of $(\frac{4 x}{5} + \frac{5}{2 x ^{2}})^{9}$ is ______.

In the binomial expansion of $(a - b)^{n}, n \geq 5$ , the sum of the $5^{t h}$ and $6^{t h}$ terms is zero. Then $a / b$ equals