JEE Main 2022 (28 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: 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 ______.

Q: Why is the sum of even binomial coefficients $2^{n-1}$?

Consider the expansion of $(1+x)^n$ and $(1-x)^n$. If you add them, the odd terms cancel out, leaving $2 \times (\text{sum of even terms})$. Specifically, $(1+1)^n + (1-1)^n = 2^n + 0 = 2 \times (\sum_{k \text{ even}} {}^{n}C_k)$. Dividing by 2 gives $2^{n-1}$.

Q: Why did we subtract ${}^{50}C_0$ in the second bracket?

The property $\sum_{k \text{ even}} {}^{n}C_k = 2^{n-1}$ includes the term for $r=0$ (i.e., ${}^{n}C_0$). Our series $S_2$ started at $r=2$. Since ${}^{50}C_0 = 1$, we had to subtract it from the total sum to match our specific series.

Enter Numerical Value:

Visualized Solution

Analyze the Expression Structure

Given expression: $1 + (^{49} C_{0} +^{49} C_{1} + \dots +^{49} C_{49} + 1) (^{50} C_{2} +^{50} C_{4} + \dots +^{50} C_{50})$
We need to find $n + m$ where the expression equals $2^{n} \cdot m$ and $m$ is odd.
Let's define the first part as $S_{1}$ and the second part as $S_{2}$ .

Defining the First Sum $S_{1}$

Let $S_{1} =^{49} C_{0} +^{49} C_{1} + \dots +^{49} C_{49}$
This is the sum of all binomial coefficients for $n = 49$ .

Applying the Sum Property $\sum^{n} C_{r} = 2^{n}$

Using the property: $\sum_{r = 0}^{n}^{n} C_{r} = 2^{n}$
For $n = 49$ : $S_{1} = 2^{49}$
The first bracket becomes: $(S_{1} + 1) = (2^{49} + 1)$

Analyzing the Second Sum $S_{2}$

Let $S_{2} =^{50} C_{2} +^{50} C_{4} + \dots +^{50} C_{50}$
This is a sum of even binomial coefficients for $n = 50$ , but it starts from $r = 2$ instead of $r = 0$ .

Sum of Even Coefficients Property

Property: $^{n} C_{0} +^{n} C_{2} +^{n} C_{4} + \dots = 2^{n - 1}$
For $n = 50$ : $^{50} C_{0} +^{50} C_{2} +^{50} C_{4} + \dots +^{50} C_{50} = 2^{50 - 1} = 2^{49}$

Adjusting for the Missing Term

We have: $S_{2} = (^{50} C_{0} +^{50} C_{2} + \dots +^{50} C_{50}) -^{50} C_{0}$
Since $^{50} C_{0} = 1$ , then $S_{2} = 2^{49} - 1$ .

Substituting Back into the Expression

Substitute $S_{1} + 1$ and $S_{2}$ into the expression:
$1 + (2^{49} + 1) (2^{49} - 1)$

Using the Difference of Squares Identity

Using identity: $(a + b) (a - b) = a^{2} - b^{2}$
Here $a = 2^{49}$ and $b = 1$ .
So, $(2^{49} + 1) (2^{49} - 1) = (2^{49})^{2} - 1^{2}$

Simplifying the Product

$(2^{49})^{2} - 1^{2} = 2^{49 \times 2} - 1 = 2^{98} - 1$

Final Simplification of the Expression

Expression = $1 + (2^{98} - 1)$
Expression = $2^{98}$

Comparing with $2^{n} \cdot m$

Compare $2^{98}$ with $2^{n} \cdot m$ .
We can write $2^{98} = 2^{98} \cdot 1$ .
Since $1$ is odd, we have $n = 98$ and $m = 1$ .

Calculating $n + m$

Calculate the final sum: $n + m$
$n + m = 98 + 1 = 99$
Final Answer: 99

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The given expression is:

1 + (^{49} C_{0} +^{49} C_{1} + \dots +^{49} C_{49} + 1) (^{50} C_{2} +^{50} C_{4} + \dots +^{50} C_{50})

This expression appears complex, but it can be decomposed into two distinct binomial sums that follow standard combinatorial identities.

Phase 1

The First Bracket - The Power of Sums

Let the first part be

S_{1}

. The expression inside the first bracket is the sum of all binomial coefficients for

n = 49

r = 0 \sum 49^{49} C_{r} = 2^{49}

The bracket in our expression is actually

S_{1} + 1

, which evaluates to:

2^{49} + 1

Phase 2

The Second Bracket - The Even Trap

Now, consider the second bracket,

S_{2} =^{50} C_{2} +^{50} C_{4} + \dots +^{50} C_{50}

. This is a sum of even binomial coefficients for

n = 50

The standard property for the sum of even coefficients is

\sum_{k even}^{n} C_{k} = 2^{n - 1}

. For

n = 50

, the sum from

r = 0

r = 50

2^{50 - 1} = 2^{49}

However, our series

S_{2}

starts at

r = 2

, omitting the term

^{50} C_{0}

. Since

^{50} C_{0} = 1

, we subtract it from the total sum:

S_{2} = 2^{49} - 1

Phase 3

The Algebraic Climax

We now substitute these values back into the original expression:

1 + (2^{49} + 1) (2^{49} - 1)

Recognizing the difference of squares identity,

(a + b) (a - b) = a^{2} - b^{2}

, where

a = 2^{49}

and

b = 1

1 + ((2^{49})^{2} - 1^{2}) = 1 + 2^{98} - 1

The constants cancel out, leaving us with the simplified result:

2^{98}

Final Calculation

The problem states that the expression equals

2^{n} \cdot m

, where

m

is odd. We have:

2^{98} = 2^{98} \cdot 1

Here,

n = 98

and

m = 1

. Since

1

is an odd number, the conditions are satisfied.

The final value is

n + m = 98 + 1 = 99

Common Traps & FAQs

Wait, why is the sum of all binomial coefficients equal to

2^{n}

This is a fundamental result derived from the Binomial Theorem. If you expand

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

and set

x = 1

, you get

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

, which simplifies to

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

. It is a beautiful, direct consequence of the theorem!

Why is the sum of even binomial coefficients

2^{n - 1}

Consider the expansion of

(1 + x)^{n}

and

(1 - x)^{n}

. If you add them, the odd terms cancel out, leaving

2 \times (sum of even terms)

. Specifically,

(1 + 1)^{n} + (1 - 1)^{n} = 2^{n} + 0 = 2 \times (\sum_{k even}^{n} C_{k})

. Dividing by 2 gives

2^{n - 1}

Why did we subtract

^{50} C_{0}

in the second bracket?

The property

\sum_{k even}^{n} C_{k} = 2^{n - 1}

includes the term for

r = 0

(i.e.,

^{n} C_{0}

). Our series

S_{2}

started at

r = 2

. Since

^{50} C_{0} = 1

, we had to subtract it from the total sum to match our specific series.

m = 1

considered an odd number?

Yes, absolutely! An odd number is any integer that is not divisible by 2. Since 1 is not divisible by 2, it is definitely odd. The condition

m

is odd is satisfied.

Animated Solution for Mathematics - Binomial Theorem: 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 ______.

Enter Numerical Value:

Visualized Solution

Analyze the Expression Structure

Defining the First Sum $S_{1}$

Applying the Sum Property $\sum^{n} C_{r} = 2^{n}$

Analyzing the Second Sum $S_{2}$

Sum of Even Coefficients Property

Adjusting for the Missing Term

Substituting Back into the Expression

Using the Difference of Squares Identity

Simplifying the Product

Final Simplification of the Expression

Comparing with $2^{n} \cdot m$

Calculating $n + m$

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

Similar Questions

If $(^{40} C_{0}) + (^{41} C_{1}) + (^{42} C_{2}) + \dots + (^{60} C_{20}) = \frac{m}{n}^{60} C_{20}$ , $m$ and $n$ are coprime, then $m + n$ 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:

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 ___

The value of $\frac{1}{1 ! 50 !} + \frac{1}{3 ! 48 !} + \frac{1}{5 ! 46 !} + \dots + \frac{1}{49 ! 2 !} + \frac{1}{51 ! 1 !}$ is

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

Let $m, n \in N$ and gcd $(2, n) = 1$ . If $30 (0 30) + 29 (1 30) + .... + 2 (28 30) + 1 (29 30) = n . 2^{m}$ , then $n + m$ is equal to

The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + \dots + x^{1000}$ is :

If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \dots + 101 \cdot C_{25} = 2^{25} \cdot k$ then $k$ is equal to__________.

If $1^{2} \cdot (^{15} C_{1}) + 2^{2} \cdot (^{15} C_{2}) + 3^{2} \cdot (^{15} C_{3}) + \dots + 1 5^{2} \cdot (^{15} C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where $m, n, k \in N$ , then $m + n + k$ is equal to :

Enter Numerical Value:

Visualized Solution

Analyze the Expression Structure

Defining the First Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Applying the Sum Property .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑nCr​=2n

Analyzing the Second Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

Sum of Even Coefficients Property

Adjusting for the Missing Term

Substituting Back into the Expression

Using the Difference of Squares Identity

Simplifying the Product

Final Simplification of the Expression

Comparing with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2n⋅m

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

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of 1!50!1​+3!48!1​+5!46!1​+⋯+49!2!1​+51!1!1​ is

.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; } Let m,n∈N and gcd (2,n)=1. If 30(030​)+29(130​)+....+2(2830​)+1(2930​)=n.2m, then n+m is equal to

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If Cr​=25Cr​ and C0​+5⋅C1​+9⋅C2​+⋯+101⋅C25​=225⋅k then k is equal to__________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 12⋅(15C1​)+22⋅(15C2​)+32⋅(15C3​)+⋯+152⋅(15C15​)=2m⋅3n⋅5k, where m,n,k∈N, then m+n+k is equal to :

Defining the First Sum $S_{1}$

Applying the Sum Property $\sum^{n} C_{r} = 2^{n}$

Analyzing the Second Sum $S_{2}$

Comparing with $2^{n} \cdot m$

Calculating $n + m$

If $(^{40} C_{0}) + (^{41} C_{1}) + (^{42} C_{2}) + \dots + (^{60} C_{20}) = \frac{m}{n}^{60} C_{20}$ , $m$ and $n$ are coprime, then $m + n$ is equal to ____.

The value of $\frac{1}{1 ! 50 !} + \frac{1}{3 ! 48 !} + \frac{1}{5 ! 46 !} + \dots + \frac{1}{49 ! 2 !} + \frac{1}{51 ! 1 !}$ is

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

Let $m, n \in N$ and gcd $(2, n) = 1$ . If $30 (0 30) + 29 (1 30) + .... + 2 (28 30) + 1 (29 30) = n . 2^{m}$ , then $n + m$ is equal to

The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + \dots + x^{1000}$ is :

If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \dots + 101 \cdot C_{25} = 2^{25} \cdot k$ then $k$ is equal to__________.

If $1^{2} \cdot (^{15} C_{1}) + 2^{2} \cdot (^{15} C_{2}) + 3^{2} \cdot (^{15} C_{3}) + \dots + 1 5^{2} \cdot (^{15} C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where $m, n, k \in N$ , then $m + n + k$ is equal to :