JEE Advanced 2010

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: For $r = 0, 1, ..., 10$ , let $A_{r}, B_{r}$ and $C_{r}$ denote, respectively, the coefficient of $x^{r}$ in the expansions of $(1 + x)^{10}, (1 + x)^{20}$ and $(1 + x)^{30}$ . Then $\sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})$ is equal to

Q: Why do we need to add the $r=0$ term and then subtract it? Can't we just solve the sum from $r=1$ directly?

That is a fantastic question! The reason we add the $r=0$ term is that standard combinatorial identities like Vandermonde's Identity are defined for the full range of indices, typically starting from $r=0$. By including the $r=0$ term, we can use these powerful tools directly. Once we have the result for the full sum, we simply subtract the $r=0$ term to get the sum starting from $r=1$. It is a common technique to 'complete' the sum to make it easier to handle!

Q: I'm confused about the symmetry property. Why is $\binom{20}{r}$ the same as $\binom{20}{20-r}$? Does this always work?

Yes, it absolutely does! The symmetry property $\binom{n}{r} = \binom{n}{n-r}$ is a fundamental property of binomial coefficients. It essentially says that choosing $r$ items out of $n$ is the same as choosing the $n-r$ items that you leave behind. It is a beautiful geometric reality of combinations. We use it here to manipulate the indices so that the sum of the lower indices in our product $\binom{n}{r} \binom{m}{k-r}$ becomes a constant, which is exactly what Vandermonde's Identity requires.

Q: How do I know which polynomial to multiply to get the coefficient? Is there a trick to picking the right powers?

The trick is to look at the sum of the lower indices. In $\sum \binom{10}{r} \binom{20}{20-r}$, the lower indices $r$ and $20-r$ add up to $20$. This tells us that the sum is the coefficient of $x^{20}$ in the product of $(1+x)^{10}$ and $(1+x)^{20}$. By multiplying the polynomials, we add the exponents, giving us $(1+x)^{30}$. So, we are looking for the coefficient of $x^{20}$ in $(1+x)^{30}$, which is $\binom{30}{20}$. It is all about matching the powers!

Select Answer:

Visualized Solution

Defining $A_{r}, B_{r}, C_{r}$

Given: $A_{r} = (r 10)$
Given: $B_{r} = (r 20)$
Given: $C_{r} = (r 30)$
The range of $r$ is $0, 1, \dots, 10$ .

Expanding the Sum

Expression: $\sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})$
Distribute $A_{r}$ : $\sum_{r = 1}^{10} (B_{10} A_{r} B_{r} - C_{10} A_{r}^{2})$
Separate the sums: $B_{10} \sum_{r = 1}^{10} A_{r} B_{r} - C_{10} \sum_{r = 1}^{10} A_{r}^{2}$

Identifying $\sum A_{r} B_{r}$

Consider the sum from $r = 0$ : $\sum_{r = 0}^{10} A_{r} B_{r} = \sum_{r = 0}^{10} (r 10) (r 20)$
Apply symmetry: $(r 20) = (20 - r 20)$
Rewrite sum: $\sum_{r = 0}^{10} (r 10) (20 - r 20)$

Calculating $\sum_{r = 0}^{10} A_{r} B_{r}$

$\sum_{r = 0}^{10} (r 10) (20 - r 20)$ is the coefficient of $x^{20}$ in $(1 + x)^{10} (1 + x)^{20}$
$(1 + x)^{10} (1 + x)^{20} = (1 + x)^{30}$
Coefficient of $x^{20}$ in $(1 + x)^{30}$ is $(20 30) = (10 30)$
Thus, $\sum_{r = 0}^{10} A_{r} B_{r} = C_{10}$

Adjusting for $r = 1$

We need $\sum_{r = 1}^{10} A_{r} B_{r} = (\sum_{r = 0}^{10} A_{r} B_{r}) - A_{0} B_{0}$
Calculate $A_{0} B_{0} = (0 10) (0 20) = 1 \times 1 = 1$
Result: $\sum_{r = 1}^{10} A_{r} B_{r} = C_{10} - 1$

Identifying $\sum A_{r}^{2}$

Consider the sum from $r = 0$ : $\sum_{r = 0}^{10} A_{r}^{2} = \sum_{r = 0}^{10} (r 10) (r 10)$
Apply symmetry: $(r 10) = (10 - r 10)$
Rewrite sum: $\sum_{r = 0}^{10} (r 10) (10 - r 10)$

Calculating $\sum_{r = 0}^{10} A_{r}^{2}$

$\sum_{r = 0}^{10} (r 10) (10 - r 10)$ is the coefficient of $x^{10}$ in $(1 + x)^{10} (1 + x)^{10}$
$(1 + x)^{10} (1 + x)^{10} = (1 + x)^{20}$
Coefficient of $x^{10}$ in $(1 + x)^{20}$ is $(10 20) = B_{10}$
Thus, $\sum_{r = 0}^{10} A_{r}^{2} = B_{10}$

Adjusting for $r = 1$

We need $\sum_{r = 1}^{10} A_{r}^{2} = (\sum_{r = 0}^{10} A_{r}^{2}) - A_{0}^{2}$
Calculate $A_{0}^{2} = ((0 10))^{2} = 1^{2} = 1$
Result: $\sum_{r = 1}^{10} A_{r}^{2} = B_{10} - 1$

Final Substitution

Substitute back: $B_{10} (C_{10} - 1) - C_{10} (B_{10} - 1)$
First term: $B_{10} C_{10} - B_{10}$
Second term: $- C_{10} B_{10} + C_{10}$

Final Simplification

Expression = $B_{10} C_{10} - B_{10} - C_{10} B_{10} + C_{10}$
Cancel $B_{10} C_{10}$ and $- C_{10} B_{10}$
Remaining: $C_{10} - B_{10}$
The final answer is $C_{10} - B_{10}$

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The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

We are given the binomial coefficients

A_{r} = (r 10)

B_{r} = (r 20)

, and

C_{r} = (r 30)

. Our objective is to evaluate the sum:

S = r = 1 \sum 10 A_{r} (B_{10} B_{r} - C_{10} A_{r})

By distributing

A_{r}

inside the summation, we can decompose the expression into two distinct parts:

S = B_{10} r = 1 \sum 10 A_{r} B_{r} - C_{10} r = 1 \sum 10 A_{r}^{2}

The Vandermonde Magic

To evaluate the first sum,

\sum_{r = 1}^{10} A_{r} B_{r}

, we first consider the sum from

r = 0

10

r = 0 \sum 10 (r 10) (r 20)

Using the symmetry property

(r 20) = (20 - r 20)

, the sum becomes

\sum_{r = 0}^{10} (r 10) (20 - r 20)

. This represents the coefficient of

x^{20}

in the expansion of

(1 + x)^{10} (1 + x)^{20} = (1 + x)^{30}

, which is

(20 30) = C_{10}

Since our target sum starts at

r = 1

, we subtract the

r = 0

term, where

A_{0} B_{0} = (0 10) (0 20) = 1

. Thus:

r = 1 \sum 10 A_{r} B_{r} = C_{10} - 1

The Symmetry Trick

Next, we evaluate the second sum,

\sum_{r = 1}^{10} A_{r}^{2}

. We consider the sum from

r = 0

10

r = 0 \sum 10 (r 10) (r 10)

Applying the symmetry property

(r 10) = (10 - r 10)

, the sum becomes

\sum_{r = 0}^{10} (r 10) (10 - r 10)

. This is the coefficient of

x^{10}

(1 + x)^{10} (1 + x)^{10} = (1 + x)^{20}

, which is

(10 20) = B_{10}

Subtracting the

r = 0

term,

A_{0}^{2} = (0 10)^{2} = 1

, we obtain:

r = 1 \sum 10 A_{r}^{2} = B_{10} - 1

The Final Collapse

Now, we substitute these results back into our master equation for

S

S = B_{10} (C_{10} - 1) - C_{10} (B_{10} - 1)

Expanding the terms, we get:

S = B_{10} C_{10} - B_{10} - C_{10} B_{10} + C_{10}

The terms

B_{10} C_{10}

and

- C_{10} B_{10}

cancel out, leaving us with the final result:

$S = C_{10} - B_{10}$

Common Traps & FAQs

Why do we need to add the

r = 0

term and then subtract it? Can't we just solve the sum from

r = 1

directly?

That is a fantastic question! The reason we add the

r = 0

term is that standard combinatorial identities like Vandermonde's Identity are defined for the full range of indices, typically starting from

r = 0

. By including the

r = 0

term, we can use these powerful tools directly. Once we have the result for the full sum, we simply subtract the

r = 0

term to get the sum starting from

r = 1

. It is a common technique to 'complete' the sum to make it easier to handle!

I'm confused about the symmetry property. Why is

(r 20)

the same as

(20 - r 20)

? Does this always work?

Yes, it absolutely does! The symmetry property

(r n) = (n - r n)

is a fundamental property of binomial coefficients. It essentially says that choosing

r

items out of

n

is the same as choosing the

n - r

items that you leave behind. It is a beautiful geometric reality of combinations. We use it here to manipulate the indices so that the sum of the lower indices in our product

(r n) (k - r m)

becomes a constant, which is exactly what Vandermonde's Identity requires.

How do I know which polynomial to multiply to get the coefficient? Is there a trick to picking the right powers?

The trick is to look at the sum of the lower indices. In

\sum (r 10) (20 - r 20)

, the lower indices

r

and

20 - r

add up to

20

. This tells us that the sum is the coefficient of

x^{20}

in the product of

(1 + x)^{10}

and

(1 + x)^{20}

. By multiplying the polynomials, we add the exponents, giving us

(1 + x)^{30}

. So, we are looking for the coefficient of

x^{20}

(1 + x)^{30}

, which is

(20 30)

. It is all about matching the powers!

Select Answer:

Visualized Solution

Defining $A_{r}, B_{r}, C_{r}$

Expanding the Sum

Identifying $\sum A_{r} B_{r}$

Calculating $\sum_{r = 0}^{10} A_{r} B_{r}$

Adjusting for $r = 1$

Identifying $\sum A_{r}^{2}$

Calculating $\sum_{r = 0}^{10} A_{r}^{2}$

Adjusting for $r = 1$

Final Substitution

Final Simplification

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Vandermonde Magic

The Symmetry Trick

The Final Collapse

Similar Questions

Let $^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1 + x)^{n}$ . If $\sum_{k = 0}^{10} (2^{2} + 3 k)^{10} C_{k} = α \cdot 3^{10} + β \cdot 2^{10}, α, β \in R$ , then $α + β$ is equal to ___

If $a_{r}$ is the coefficient of $x^{10 - r}$ in the Binomial expansion of $(1 + x)^{10}$ , then $\sum_{r = 1}^{10} r^{3} (\frac{a _{r}}{a _{r - 1}})^{2}$ is equal to

If $^{20} C_{r}$ is the co-efficient of $x^{r}$ in the expansion of $(1 + x)^{20}$ , then the value of $\sum_{r = 0}^{20} r^{2}^{20} C_{r}$ is equal to :

If $A$ denotes the sum of all the coefficients in the expansion of $(1 - 3 x + 10 x^{2})^{n}$ and $B$ denotes the sum of all the coefficients in the expansion of $(1 + x^{2})^{n}$ , then :

The sum, of the coefficients of the first 50 terms in the binomial expansion of $(1 - x)^{100}$ , is equal to

Let $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}$ , then $a_{1} + a_{3} + a_{5} + \dots + a_{37}$ is equal to

The value of $(0 30) (10 30) - (1 30) (11 30) + (2 30) (12 30) - .... + (20 30) (30 30)$ is

For $2 \leq r \leq n$ , $(r n) + 2 (r - 1 n) + (r - 2 n) =$

The sum of the series $^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + .... +^{20} C_{10}$ is

For some $n \neq = 10$ , let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of $(1 + x)^{n + 4}$ be in A.P. Then the largest coefficient in the expansion of $(1 + x)^{n + 4}$ is:

Select Answer:

Visualized Solution

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ar​,Br​,Cr​

Expanding the Sum

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑Ar​Br​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑r=010​Ar​Br​

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

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑Ar2​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑r=010​Ar2​

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

Final Substitution

Final Simplification

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Vandermonde Magic

The Symmetry Trick

The Final Collapse

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let nCr​ denote the binomial coefficient of xr in the expansion of (1+x)n. If ∑k=010​(22+3k)10Ck​=α⋅310+β⋅210,α,β∈R, 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 ar​ is the coefficient of x10−r in the Binomial expansion of (1+x)10, then ∑r=110​r3(ar−1​ar​​)2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 20Cr​ is the co-efficient of xr in the expansion of (1+x)20, then the value of ∑r=020​r220Cr​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A denotes the sum of all the coefficients in the expansion of (1−3x+10x2)n and B denotes the sum of all the coefficients in the expansion of (1+x2)n, then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum, of the coefficients of the first 50 terms in the binomial expansion of (1−x)100, is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (1+x+2x2)20=a0​+a1​x+a2​x2+⋯+a40​x40, then a1​+a3​+a5​+⋯+a37​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of (030​)(1030​)−(130​)(1130​)+(230​)(1230​)−....+(2030​)(3030​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For 2≤r≤n, (rn​)+2(r−1n​)+(r−2n​)=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the series 20C0​−20C1​+20C2​−20C3​+....+20C10​ is

Defining $A_{r}, B_{r}, C_{r}$

Identifying $\sum A_{r} B_{r}$

Calculating $\sum_{r = 0}^{10} A_{r} B_{r}$

Adjusting for $r = 1$

Identifying $\sum A_{r}^{2}$

Calculating $\sum_{r = 0}^{10} A_{r}^{2}$

Adjusting for $r = 1$

Let $^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1 + x)^{n}$ . If $\sum_{k = 0}^{10} (2^{2} + 3 k)^{10} C_{k} = α \cdot 3^{10} + β \cdot 2^{10}, α, β \in R$ , then $α + β$ is equal to ___

If $a_{r}$ is the coefficient of $x^{10 - r}$ in the Binomial expansion of $(1 + x)^{10}$ , then $\sum_{r = 1}^{10} r^{3} (\frac{a _{r}}{a _{r - 1}})^{2}$ is equal to

If $^{20} C_{r}$ is the co-efficient of $x^{r}$ in the expansion of $(1 + x)^{20}$ , then the value of $\sum_{r = 0}^{20} r^{2}^{20} C_{r}$ is equal to :

If $A$ denotes the sum of all the coefficients in the expansion of $(1 - 3 x + 10 x^{2})^{n}$ and $B$ denotes the sum of all the coefficients in the expansion of $(1 + x^{2})^{n}$ , then :

The sum, of the coefficients of the first 50 terms in the binomial expansion of $(1 - x)^{100}$ , is equal to

Let $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}$ , then $a_{1} + a_{3} + a_{5} + \dots + a_{37}$ is equal to

The value of $(0 30) (10 30) - (1 30) (11 30) + (2 30) (12 30) - .... + (20 30) (30 30)$ is

For $2 \leq r \leq n$ , $(r n) + 2 (r - 1 n) + (r - 2 n) =$

The sum of the series $^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + .... +^{20} C_{10}$ is