JEE Main 2022 (26 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: 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 ____.

Q: Wait, why do we change ${}^{40}C_0$ to ${}^{40}C_{40}$? Doesn't that change the value of the series?

Great question! The symmetry property ${}^{n}C_{r} = {}^{n}C_{n-r}$ ensures that ${}^{40}C_0$ is exactly equal to ${}^{40}C_{40}$. We aren't changing the value; we are just changing the representation to reveal the hidden structure of the series.

Q: Can I use the Hockey-stick identity directly on the original series without symmetry?

No, you cannot. The Hockey-stick identity $\sum_{i=k}^{n} {}^{i}C_{k} = {}^{n+1}C_{k+1}$ requires the lower index ($k$) to be constant across all terms. In the original series, the lower index changes with every term. That is why the symmetry transformation is the crucial first step.

Q: How do I know if the Hockey-stick identity applies to my specific range?

The identity $\sum_{i=k}^{n} {}^{i}C_{k} = {}^{n+1}C_{k+1}$ is valid as long as the summation starts from $i=k$. In our case, the first term is ${}^{40}C_{40}$, which is exactly ${}^{k}C_{k}$ with $k=40$. Since it starts at the correct index, the identity applies perfectly.

Enter Numerical Value:

Progressive Hints

Hint 1

The series looks intimidating, but notice the pattern of indices. The upper index increases by 1, and the lower index also increases by 1. This is a classic setup for the symmetry property of binomial coefficients:

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

. Try applying this to every term.

Hint 2

Once you apply the symmetry property, the lower index becomes constant for all terms. Now, look at the sum of binomial coefficients where the upper index varies but the lower index remains fixed. This is the exact structure required for the Hockey-stick identity:

\sum_{i = k}^{n}^{i} C_{k} =^{n + 1} C_{k + 1}

Hint 3

After simplifying the sum using the identity, you will have a single binomial coefficient. To match the target form

\frac{m}{n}^{60} C_{20}

, use the expansion property

^{n} C_{r} = \frac{n}{r}^{n - 1} C_{r - 1}

to manipulate your result into the desired form.

Visualized Solution

The Given Series

Given series: $S =^{40} C_{0} +^{41} C_{1} +^{42} C_{2} + \dots +^{60} C_{20}$
Target form: $S = \frac{m}{n}^{60} C_{20}$
Observe the pattern: Both $n$ and $r$ increase by $1$ in each term.

Applying Symmetry Property

Property: $^{n} C_{r} =^{n} C_{n - r}$
We will apply this symmetry property to every term in the summation.

The Transformed Series

Term 1: $^{40} C_{0} =^{40} C_{40 - 0} =^{40} C_{40}$
Term 2: $^{41} C_{1} =^{41} C_{41 - 1} =^{41} C_{40}$
Term 21: $^{60} C_{20} =^{60} C_{60 - 20} =^{60} C_{40}$
Transformed Series: $S =^{40} C_{40} +^{41} C_{40} +^{42} C_{40} + \dots +^{60} C_{40}$

The Hockey-stick Identity

Hockey-stick Identity: $\sum_{i = k}^{n}^{i} C_{k} =^{n + 1} C_{k + 1}$
This identity is valid when the summation starts exactly from $i = k$ .

Applying the Identity

In our series: $k = 40$ and $n = 60$ .
The sum starts exactly at $^{k} C_{k} =^{40} C_{40}$ .
Applying the formula: $S =^{60 + 1} C_{40 + 1}$

Simplifying the Result

Sum $S =^{61} C_{41}$
We need to express this in terms of $^{60} C_{20}$ .

The Expansion Property

Expansion Property: $^{n} C_{r} = \frac{n}{r}^{n - 1} C_{r - 1}$
This allows us to extract factors and reduce the indices.

Relating to $^{60} C_{20}$

Substitute $n = 61, r = 41$ :
$S = \frac{61}{41}^{60} C_{40}$
Using symmetry again: $^{60} C_{40} =^{60} C_{20}$
$S = \frac{61}{41}^{60} C_{20}$

Identifying $m$ and $n$

Comparing with $\frac{m}{n}^{60} C_{20}$ :
$m = 61$ , $n = 41$
Check: $61$ and $41$ are both prime, so they are coprime.

Final Sum $m + n$

Final calculation: $m + n = 61 + 41$
$m + n = 102$
Final Answer: 102

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

We are tasked with evaluating the sum:

S =^{40} C_{0} +^{41} C_{1} +^{42} C_{2} + \dots +^{60} C_{20}

At first glance, this series appears complex. However, we can simplify it by utilizing the fundamental symmetry property of binomial coefficients:

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

Applying Symmetry

By applying this identity to every term in the series, we transform the expression into a more manageable form:

S =^{40} C_{40} +^{41} C_{40} +^{42} C_{40} + \dots +^{60} C_{40}

Notice that the lower index is now constant. This structure is the classic setup for the Hockey-stick identity.

The Hockey-stick Identity

The Hockey-stick identity states that:

i = k \sum n^{i} C_{k} =^{n + 1} C_{k + 1}

In our specific case, we have

k = 40

and

n = 60

. Substituting these values into the identity, the sum simplifies to:

S =^{61} C_{41}

Final Calculation

To express this in terms of the requested format, we use the property:

^{n} C_{r} = \frac{n}{r}^{n - 1} C_{r - 1}

Applying this to our result, we obtain:

S = \frac{61}{41}^{60} C_{40}

Since

^{60} C_{40} =^{60} C_{20}

, our sum is equivalent to:

S = \frac{61}{41}^{60} C_{20}

Given the form

m = 61

and

n = 41

, we find the final result: $m + n = 61 + 41 = 102$

Common Traps & FAQs

Wait, why do we change

^{40} C_{0}

^{40} C_{40}

? Doesn't that change the value of the series?

Great question! The symmetry property

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

ensures that

^{40} C_{0}

is exactly equal to

^{40} C_{40}

. We aren't changing the value; we are just changing the representation to reveal the hidden structure of the series.

Can I use the Hockey-stick identity directly on the original series without symmetry?

No, you cannot. The Hockey-stick identity

\sum_{i = k}^{n}^{i} C_{k} =^{n + 1} C_{k + 1}

requires the lower index (

k

) to be constant across all terms. In the original series, the lower index changes with every term. That is why the symmetry transformation is the crucial first step.

How do I know if the Hockey-stick identity applies to my specific range?

The identity

\sum_{i = k}^{n}^{i} C_{k} =^{n + 1} C_{k + 1}

is valid as long as the summation starts from

i = k

. In our case, the first term is

^{40} C_{40}

, which is exactly

^{k} C_{k}

with

k = 40

. Since it starts at the correct index, the identity applies perfectly.

Animated Solution for Mathematics - Binomial Theorem: 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 ____.

Enter Numerical Value:

Visualized Solution

The Given Series

Applying Symmetry Property

The Transformed Series

The Hockey-stick Identity

Applying the Identity

Simplifying the Result

The Expansion Property

Relating to $^{60} C_{20}$

Identifying $m$ and $n$

Final Sum $m + n$

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Applying Symmetry

The Hockey-stick Identity

Final Calculation

Similar Questions

If $\frac{^{11} C _{1}}{2} + \frac{^{11} C _{2}}{3} + \dots + \frac{^{11} C _{9}}{10} = \frac{n}{m}$ with $g cd (n, m) = 1$ , then $n + m$ is equal to

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 ______.

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (40C0​)+(41C1​)+(42C2​)+⋯+(60C20​)=nm​60C20​, m and n are coprime, then m+n is equal to ____.

Enter Numerical Value:

Visualized Solution

The Given Series

Applying Symmetry Property

The Transformed Series

The Hockey-stick Identity

Applying the Identity

Simplifying the Result

The Expansion Property

Relating to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 60C20​

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

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

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Applying Symmetry

The Hockey-stick Identity

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Binomial Theorem: 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 ____.

Relating to $^{60} C_{20}$

Identifying $m$ and $n$

Final Sum $m + n$

If $\frac{^{11} C _{1}}{2} + \frac{^{11} C _{2}}{3} + \dots + \frac{^{11} C _{9}}{10} = \frac{n}{m}$ with $g cd (n, m) = 1$ , then $n + m$ is equal to

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 ______.

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

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 :

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 sum of the series $2 .^{20} C_{0} + 5 .^{20} C_{1} + 8 .^{20} C_{2} + 11 .^{20} C_{3} + ... + 62 .^{20} C_{20}$ is equal to :