JEE Main 2022 (29 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: If $\sum_{k = 1}^{10} k^{2} (^{10} C_{k})^{2} = 22000 L$ , then $L$ is equal to ______.

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Look closely at the term

k^{2} (^{10} C_{k})^{2}

. The variable

k

is the intruder preventing us from using standard binomial identities. Your goal is to 'absorb' this

k

into the binomial coefficient itself. How can you rewrite

k \cdot^{10} C_{k}

to make it look like a simpler combination?

Hint 2

The Arsenal: You need the 'Absorption Property' of binomial coefficients. Recall that

k \cdot^{n} C_{k} = n \cdot^{n - 1} C_{k - 1}

. This is the surgical tool that will allow you to strip away the

k

and transform the expression into a form where the summation identity

\sum (^{n} C_{r})^{2} =^{2 n} C_{n}

can be applied.

Hint 3

The Execution: Start by rewriting the sum as

\sum_{k = 1}^{10} (k \cdot^{10} C_{k})^{2}

. Apply the absorption property to the inner term to get

10 \cdot^{9} C_{k - 1}

. Now, square this term to get

100 \cdot (^{9} C_{k - 1})^{2}

. Finally, shift your index

r = k - 1

to transform the sum into

100 \sum_{r = 0}^{9} (^{9} C_{r})^{2}

Visualized Solution

Analyze the Given Summation

Given equation: $\sum_{k = 1}^{10} k^{2} (^{10} C_{k})^{2} = 22000 L$
Objective: Find the value of the constant $L$ .
The general term of the sum is $(k \cdot^{10} C_{k})^{2}$ .

Focus on the General Term

Rewrite the general term: $k^{2} (^{10} C_{k})^{2} = (k \cdot^{10} C_{k})^{2}$
We need to simplify the expression $k \cdot^{10} C_{k}$ inside the square.

Apply the Absorption Property

Using the property: $k \cdot^{n} C_{k} = n \cdot^{n - 1} C_{k - 1}$
For $n = 10$ , we get: $k \cdot^{10} C_{k} = 10 \cdot^{9} C_{k - 1}$

Square the Simplified Term

Substitute the simplified term back: $(k \cdot^{10} C_{k})^{2} = (10 \cdot^{9} C_{k - 1})^{2}$
Expanding the square: $100 \cdot (^{9} C_{k - 1})^{2}$

Restructure the Summation

The sum becomes: $100 \sum_{k = 1}^{10} (^{9} C_{k - 1})^{2}$
Let $r = k - 1$ . As $k$ goes from $1$ to $10$ , $r$ goes from $0$ to $9$ .
The sum is now: $100 \sum_{r = 0}^{9} (^{9} C_{r})^{2}$

Identify the Square Sum Identity

Standard Identity: $\sum_{r = 0}^{n} (^{n} C_{r})^{2} =^{2 n} C_{n}$
Applying for $n = 9$ : $\sum_{r = 0}^{9} (^{9} C_{r})^{2} =^{18} C_{9}$

Calculate the Total Sum

The total sum is: $100 \cdot^{18} C_{9}$
We need to evaluate $^{18} C_{9} = \frac{18 !}{9 ! \cdot 9 !}$

Evaluate $^{18} C_{9}$ Numerically

$^{18} C_{9} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$
After simplification: $^{18} C_{9} = 48620$

Final Value of the Sum

Sum = $100 \times 48620$
Sum = $4862000$

Equate and Solve for $L$

Given: $22000 L = 4862000$
Dividing by $1000$ on both sides: $22 L = 4862$
Solve for $L$ : $L = \frac{4862}{22}$

The Final Result

$L = 221$
Final Answer: $L = 221$

Summary and Key Takeaways

Key Takeaway 1: Use $k \cdot^{n} C_{k} = n \cdot^{n - 1} C_{k - 1}$ to absorb variables.
Key Takeaway 2: Remember the identity $\sum (^{n} C_{r})^{2} =^{2 n} C_{n}$ .
Next Challenge: Try solving the same problem if the term was $k^{3} (^{10} C_{k})^{2}$ .

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

We are tasked with solving the equation:

k = 1 \sum 10 k^{2} (^{10} C_{k})^{2} = 22000 L

Many students attempt to calculate these terms individually, which leads to arithmetic errors. Instead, we focus on the core term

k^{2} (^{10} C_{k})^{2}

, which can be rewritten as

(k \cdot^{10} C_{k})^{2}

The Absorption Property

To simplify the expression, we utilize the absorption property of binomial coefficients:

k \cdot^{n} C_{k} = n \cdot^{n - 1} C_{k - 1}

Applying this property with

n = 10

, the term

k \cdot^{10} C_{k}

transforms into

10 \cdot^{9} C_{k - 1}

. This effectively absorbs the variable

k

and simplifies the expression significantly.

Restructuring the Summation

Substituting this back into our squared expression, we get:

(k \cdot^{10} C_{k})^{2} = (10 \cdot^{9} C_{k - 1})^{2} = 100 \cdot (^{9} C_{k - 1})^{2}

We can now pull the constant

100

out of the summation:

100 k = 1 \sum 10 (^{9} C_{k - 1})^{2}

By substituting

r = k - 1

, where

r

ranges from

0

9

, the expression becomes:

100 r = 0 \sum 9 (^{9} C_{r})^{2}

Applying the Identity

We use the standard identity for the sum of squares of binomial coefficients:

r = 0 \sum n (^{n} C_{r})^{2} =^{2 n} C_{n}

Setting

n = 9

, the summation simplifies to

^{18} C_{9}

. Thus, the entire left-hand side of our original equation is:

100 \cdot^{18} C_{9}

Final Calculation

Evaluating the combination

^{18} C_{9}

^{18} C_{9} = \frac{18 !}{9 ! \cdot 9 !} = 48620

Multiplying this by

100

, we find the total sum:

100 \times 48620 = 4862000

Equating this to the right-hand side of our original equation:

22000 L = 4862000

Dividing both sides by

22000

L = \frac{4862000}{22000} = \frac{4862}{22} = 221

The final value is $L = 221$ .

Common Traps & FAQs

Why can't I just expand the binomial coefficients and calculate them one by one? It is only 10 terms.

My dear student, while calculating 10 terms is technically possible, it is a trap designed to waste your precious time in the JEE Advanced exam. The goal is to develop mathematical intuition. By using the absorption property, we reduce a massive arithmetic task to a single, elegant identity. Always look for the pattern, not the brute force.

Why does the summation index change from

k = 1

10

r = 0

9

This is a change of variables, a fundamental technique in calculus and series. When we define

r = k - 1

, we are simply shifting our perspective. When

k = 1

r = 0

. When

k = 10

r = 9

. It is the same set of terms, just indexed differently to match the standard form of the identity

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

Is there a general formula for

\sum k^{2} (^{n} C_{k})^{2}

Yes, absolutely! If you follow the logic we used here, you can derive that

\sum_{k = 1}^{n} k^{2} (^{n} C_{k})^{2} = n^{2} \cdot^{2 n - 2} C_{n - 1}

. Memorizing this is helpful, but understanding the derivation—the 'why'—is what makes you a true problem solver.

Animated Solution for Mathematics - Binomial Theorem: If $\sum_{k = 1}^{10} k^{2} (^{10} C_{k})^{2} = 22000 L$ , then $L$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Analyze the Given Summation

Focus on the General Term

Apply the Absorption Property

Square the Simplified Term

Restructure the Summation

Identify the Square Sum Identity

Calculate the Total Sum

Evaluate $^{18} C_{9}$ Numerically

Final Value of the Sum

Equate and Solve for $L$

The Final Result

Summary and Key Takeaways

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Absorption Property

Restructuring the Summation

Applying the Identity

Final Calculation

Similar Questions

$\sum_{k = 0}^{20} (^{20} C_{k})^{2}$ is equal to :

$\sum_{i = 1}^{20} \frac{^{20} C _{i - 1}}{^{20} C _{i} + ^{20} C _{i - 1}} = \frac{k}{21}$ , then k equals :

Suppose $\sum_{r = 0}^{2023} r^{2} \cdot^{2023} C_{r} = 2023 \times α \times 2^{2022}$ . Then the value of $α$ 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 $\sum_{r = 0}^{25} {^{50} C_{r} \cdot^{50 - r} C_{25 - r}} = K (^{50} C_{25})$ , then K is equal to :

If $\sum_{r = 1}^{30} \frac{r ^{2} ( ^{30} C _{r} ) ^{2}}{^{30} C _{r - 1}} = α \times 2^{29}$ , then $α$ is equal to

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

The value $\sum_{r = 0}^{22}^{22} C_{r} \cdot^{23} C_{r}$ is

$\sum_{i, j = 0 i \neq = j}^{n}^{n} C_{i}^{n} C_{j}$ is equal to

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 :

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 ∑k=110​k2(10Ck​)2=22000L, then L is equal to ______.

Enter Numerical Value:

Visualized Solution

Analyze the Given Summation

Focus on the General Term

Apply the Absorption Property

Square the Simplified Term

Restructure the Summation

Identify the Square Sum Identity

Calculate the Total Sum

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 18C9​ Numerically

Final Value of the Sum

Equate and Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

The Final Result

Summary and Key Takeaways

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Absorption Property

Restructuring the Summation

Applying the Identity

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Binomial Theorem: If $\sum_{k = 1}^{10} k^{2} (^{10} C_{k})^{2} = 22000 L$ , then $L$ is equal to ______.

Evaluate $^{18} C_{9}$ Numerically

Equate and Solve for $L$

$\sum_{k = 0}^{20} (^{20} C_{k})^{2}$ is equal to :

$\sum_{i = 1}^{20} \frac{^{20} C _{i - 1}}{^{20} C _{i} + ^{20} C _{i - 1}} = \frac{k}{21}$ , then k equals :

Suppose $\sum_{r = 0}^{2023} r^{2} \cdot^{2023} C_{r} = 2023 \times α \times 2^{2022}$ . Then the value of $α$ 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 $\sum_{r = 0}^{25} {^{50} C_{r} \cdot^{50 - r} C_{25 - r}} = K (^{50} C_{25})$ , then K is equal to :

If $\sum_{r = 1}^{30} \frac{r ^{2} ( ^{30} C _{r} ) ^{2}}{^{30} C _{r - 1}} = α \times 2^{29}$ , then $α$ is equal to

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

The value $\sum_{r = 0}^{22}^{22} C_{r} \cdot^{23} C_{r}$ is

$\sum_{i, j = 0 i \neq = j}^{n}^{n} C_{i}^{n} C_{j}$ is equal to

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 :