JEE Main 2021 (20 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: For the natural numbers $m, n$ , if $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + \dots + a_{m + n} y^{m + n}$ and $a_{1} = a_{2} = 10$ , then the value of $(m + n)$ is equal to:

Q: I got confused with the signs in $(1-y)^m$. Why is the second term negative?

Think of it as $(1 + (-y))^m$. When you apply the binomial theorem, the first term is $1$, the second is $m(-y)$, which is $-my$, and the third is $\frac{m(m-1)}{2}(-y)^2$, which is positive because $(-y)^2 = y^2$. The signs alternate because of the odd powers of $-y$. Just keep track of those signs, and you will be perfectly fine!

Q: Do I need to find the individual values of $m$ and $n$ to solve this?

Not necessarily! Notice how the algebra collapses into $(n-m)^2$ and $(n+m)$. Sometimes, the structure of the equation is more important than the individual variables. In this problem, we are looking for the sum $(m+n)$, and the algebraic manipulation allows us to find that sum directly without solving for $m$ and $n$ separately. Always look for patterns before diving into solving for unknowns.

Select Answer:

Progressive Hints

Hint 1

The Spark: Do not be intimidated by the full expansion of the binomials. In JEE Advanced problems, efficiency is key. You are only given information about the coefficients of

y

and

y^{2}

. Does it make sense to expand the entire polynomial up to

y^{m + n}

? Focus your energy only on the terms that contribute to

y

and

y^{2}

Hint 2

The Arsenal: Recall the generalized Binomial Theorem:

(1 + x)^{k} = 1 + k x + \frac{k ( k - 1 )}{2} x^{2} + \dots

. Apply this expansion to both

(1 - y)^{m}

and

(1 + y)^{n}

. Be extremely careful with the signs: in the first bracket, your

x

- y

, and in the second, it is

+ y

Hint 3

The Execution: Once you have the expansions, multiply them and collect the coefficients. You will find

a_{1} = n - m = 10

. Now, express

a_{2}

in terms of

m

and

n

. Can you see a way to manipulate the algebraic expression for

a_{2}

to reveal the structure of

(n - m)^{2}

and

(n + m)

? This is where the algebra simplifies beautifully.

Visualized Solution

Introduction to the Problem

Given: $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + \dots + a_{m + n} y^{m + n}$
Given: $a_{1} = 10$ and $a_{2} = 10$
Objective: Find the value of $(m + n)$

Expanding $(1 - y)^{m}$

Recall Binomial Theorem: $(1 + x)^{k} = 1 + k x + \frac{k ( k - 1 )}{2} x^{2} + \dots$
Expanding the first term:
$(1 - y)^{m} = 1 - m y + \frac{m ( m - 1 )}{2} y^{2} - \dots$

Expanding $(1 + y)^{n}$

Expanding the second term:
$(1 + y)^{n} = 1 + n y + \frac{n ( n - 1 )}{2} y^{2} + \dots$

Finding Coefficient $a_{1}$

Product: $(1 - m y + \frac{m ( m - 1 )}{2} y^{2}) (1 + n y + \frac{n ( n - 1 )}{2} y^{2})$
To find $a_{1}$ (coefficient of $y$ ), multiply terms that result in $y$ :
$a_{1} = (1) (n y) + (- m y) (1)$
$a_{1} = n - m$

Setting up Equation 1

We found: $a_{1} = n - m$
Given in problem: $a_{1} = 10$
Therefore: $n - m = 10$ ... (Equation 1)

Finding Coefficient $a_{2}$

To find $a_{2}$ (coefficient of $y^{2}$ ), multiply terms that result in $y^{2}$ :
$1 \times \frac{n ( n - 1 )}{2} y^{2}$
$(- m y) \times (n y)$
$\frac{m ( m - 1 )}{2} y^{2} \times 1$
$a_{2} = \frac{n ( n - 1 )}{2} - mn + \frac{m ( m - 1 )}{2}$

Setting up Equation 2

We found: $a_{2} = \frac{n ( n - 1 )}{2} - mn + \frac{m ( m - 1 )}{2}$
Given in problem: $a_{2} = 10$
Therefore: $\frac{n ( n - 1 )}{2} - mn + \frac{m ( m - 1 )}{2} = 10$

Simplifying Equation 2

Multiply the entire equation by $2$ to remove fractions:
$n (n - 1) - 2 mn + m (m - 1) = 20$
Expand the brackets:
$n^{2} - n - 2 mn + m^{2} - m = 20$

Rearranging and Using Identities

Rearrange the terms strategically:
$(n^{2} + m^{2} - 2 mn) - (n + m) = 20$
Recognize the perfect square identity: $n^{2} + m^{2} - 2 mn = (n - m)^{2}$
Substitute the identity:
$(n - m)^{2} - (n + m) = 20$

Final Substitution

Recall Equation 1: $n - m = 10$
Substitute $n - m = 10$ into our simplified equation:
$(10)^{2} - (n + m) = 20$
$100 - (n + m) = 20$

Calculating the Final Answer

Solve for $(m + n)$ :
$n + m = 100 - 20$
$m + n = 80$
Final Answer: 80

00:00 / 00:00

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

To solve for the sum

(m + n)

, we begin by expanding the expression

(1 - y)^{m} (1 + y)^{n}

using the Binomial Theorem. We are interested in the coefficients

a_{1}

and

a_{2}

corresponding to

y

and

y^{2}

, respectively.

The expansions are:

(1 - y)^{m} = 1 - m y + \frac{m ( m - 1 )}{2} y^{2} - \dots

(1 + y)^{n} = 1 + n y + \frac{n ( n - 1 )}{2} y^{2} + \dots

Extracting the Coefficients

When we multiply these two series, we focus only on the terms up to

y^{2}

. The coefficient of

y

, denoted as

a_{1}

, is obtained by combining the linear terms:

a_{1} = n - m = 10

For the coefficient of

y^{2}

, denoted as

a_{2}

, we consider the three possible combinations that result in a

y^{2}

term:

a_{2} = \frac{n ( n - 1 )}{2} - mn + \frac{m ( m - 1 )}{2} = 10

The Master Equation

To simplify the expression for

a_{2}

, we multiply the entire equation by

2

to clear the denominators:

n (n - 1) - 2 mn + m (m - 1) = 20

n^{2} - n - 2 mn + m^{2} - m = 20

We can rearrange these terms to reveal a recognizable algebraic structure:

(n^{2} + m^{2} - 2 mn) - (n + m) = 20

Final Calculation

Recognizing the perfect square identity, we substitute

(n - m)^{2}

for

(n^{2} + m^{2} - 2 mn)

(n - m)^{2} - (n + m) = 20

Given that

n - m = 10

, we substitute this value into the equation:

1 0^{2} - (n + m) = 20

100 - (n + m) = 20

Solving for the sum, we find: $n + m = 80$

Common Traps & FAQs

Why do we stop at

y^{2}

? What if the problem asked for

a_{3}

That is a brilliant question! In the JEE exam, time is your most precious currency. We only expand as far as the problem demands. If the problem had asked for

a_{3}

, we would simply extend our expansion to the

y^{3}

term. Always analyze the constraints first; if the information given only involves

a_{1}

and

a_{2}

, expanding further is just extra work that increases the probability of a calculation error.

I got confused with the signs in

(1 - y)^{m}

. Why is the second term negative?

Think of it as

(1 + (- y))^{m}

. When you apply the binomial theorem, the first term is

1

, the second is

m (- y)

, which is

- m y

, and the third is

\frac{m ( m - 1 )}{2} (- y)^{2}

, which is positive because

(- y)^{2} = y^{2}

. The signs alternate because of the odd powers of

- y

. Just keep track of those signs, and you will be perfectly fine!

Do I need to find the individual values of

m

and

n

to solve this?

Not necessarily! Notice how the algebra collapses into

(n - m)^{2}

and

(n + m)

. Sometimes, the structure of the equation is more important than the individual variables. In this problem, we are looking for the sum

(m + n)

, and the algebraic manipulation allows us to find that sum directly without solving for

m

and

n

separately. Always look for patterns before diving into solving for unknowns.

Animated Solution for Mathematics - Binomial Theorem: For the natural numbers $m, n$ , if $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + \dots + a_{m + n} y^{m + n}$ and $a_{1} = a_{2} = 10$ , then the value of $(m + n)$ is equal to:

Select Answer:

Visualized Solution

Introduction to the Problem

Expanding $(1 - y)^{m}$

Expanding $(1 + y)^{n}$

Finding Coefficient $a_{1}$

Setting up Equation 1

Finding Coefficient $a_{2}$

Setting up Equation 2

Simplifying Equation 2

Rearranging and Using Identities

Final Substitution

Calculating the Final Answer

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

Extracting the Coefficients

The Master Equation

Final Calculation

Similar Questions

For natural numbers $m, n$ if $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + ....$ and $a_{1} = a_{2} = 10$ , then $(m, n)$ is

Let $n \geq 5$ be an integer. If $9^{n} - 8 n - 1 = 64 α$ and $6^{n} - 5 n - 1 = 25 β$ , then $α - β$ is equal to:

If the coefficients of $x$ and $x^{2}$ in $(1 + x)^{p} (1 - x)^{q}$ are 4 and $- 5$ respectively, then $2 p + 3 q$ is equal to

The total number of two digit numbers 'n', such that $3^{n} + 7^{n}$ is a multiple of 10, is

If in the expansion of $(1 + x)^{m} (1 - x)^{n}$ , the coefficients of $x$ and $x^{2}$ are 3 and $- 6$ respectively, then $m$ is

The number of elements in the set ${n \in {1, 2, 3, \dots, 100} : (11)^{n} > (10)^{n} + (9)^{n}}$ is ____.

The remainder when $(11)^{1011} + (1011)^{11}$ is divided by 9 is

Let the number $(22)^{2022} + (2022)^{22}$ leave the remainder $α$ when divided by 3 and $β$ when divided by 7. Then $(α^{2} + β^{2})$ is equal to

The coefficient of $x^{7}$ in the expression $(1 + x)^{10} + x (1 + x)^{9} + x^{2} (1 + x)^{8} + ... + x^{10}$ is :

The coefficient of $x^{7}$ in the expression $(1 + x)^{10} + x (1 + x)^{9} + x^{2} (1 + x)^{8} + \dots + x^{10}$ is :

Select Answer:

Visualized Solution

Introduction to the Problem

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1−y)m

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1+y)n

Finding Coefficient .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​

Setting up Equation 1

Finding Coefficient .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2​

Setting up Equation 2

Simplifying Equation 2

Rearranging and Using Identities

Final Substitution

Calculating the Final Answer

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

Extracting the Coefficients

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For natural numbers m,n if (1−y)m(1+y)n=1+a1​y+a2​y2+.... and a1​=a2​=10, then (m,n) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let n≥5 be an integer. If 9n−8n−1=64α and 6n−5n−1=25β, then α−β is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the coefficients of x and x2 in (1+x)p(1−x)q are 4 and −5 respectively, then 2p+3q is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The total number of two digit numbers 'n', such that 3n+7n is a multiple of 10, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If in the expansion of (1+x)m(1−x)n, the coefficients of x and x2 are 3 and −6 respectively, then m is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of elements in the set {n∈{1,2,3,…,100}:(11)n>(10)n+(9)n} is ____.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The remainder when (11)1011+(1011)11 is divided by 9 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the number (22)2022+(2022)22 leave the remainder α when divided by 3 and β when divided by 7. Then (α2+β2) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x7 in the expression (1+x)10+x(1+x)9+x2(1+x)8+...+x10 is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x7 in the expression (1+x)10+x(1+x)9+x2(1+x)8+⋯+x10 is :

Expanding $(1 - y)^{m}$

Expanding $(1 + y)^{n}$

Finding Coefficient $a_{1}$

Finding Coefficient $a_{2}$

For natural numbers $m, n$ if $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + ....$ and $a_{1} = a_{2} = 10$ , then $(m, n)$ is

Let $n \geq 5$ be an integer. If $9^{n} - 8 n - 1 = 64 α$ and $6^{n} - 5 n - 1 = 25 β$ , then $α - β$ is equal to:

If the coefficients of $x$ and $x^{2}$ in $(1 + x)^{p} (1 - x)^{q}$ are 4 and $- 5$ respectively, then $2 p + 3 q$ is equal to

The total number of two digit numbers 'n', such that $3^{n} + 7^{n}$ is a multiple of 10, is

If in the expansion of $(1 + x)^{m} (1 - x)^{n}$ , the coefficients of $x$ and $x^{2}$ are 3 and $- 6$ respectively, then $m$ is

The number of elements in the set ${n \in {1, 2, 3, \dots, 100} : (11)^{n} > (10)^{n} + (9)^{n}}$ is ____.

The remainder when $(11)^{1011} + (1011)^{11}$ is divided by 9 is

Let the number $(22)^{2022} + (2022)^{22}$ leave the remainder $α$ when divided by 3 and $β$ when divided by 7. Then $(α^{2} + β^{2})$ is equal to

The coefficient of $x^{7}$ in the expression $(1 + x)^{10} + x (1 + x)^{9} + x^{2} (1 + x)^{8} + ... + x^{10}$ is :

The coefficient of $x^{7}$ in the expression $(1 + x)^{10} + x (1 + x)^{9} + x^{2} (1 + x)^{8} + \dots + x^{10}$ is :