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 : Options: (A) $420 \times 2^{19}$ (B) $380 \times 2^{19}$ (C) $380 \times 2^{18}$ (D) $420 \times 2^{18}$

### Analyzing the Setup The problem asks us to evaluate the sum $S = \sum_{r=0}^{20} r^2 {}^{20}C_r$. The presence of the $r^2$ term prevents the direct application of the standard binomial identity $\sum {}^{n}C_r = 2^n$. We must transform the expression to isolate the binomial coefficients. ### The Algebraic Surgery The core difficulty is the $r^2$ term. We recall that the binomial coefficient is defined as: $$ {}^{n}C_r = \frac{n!}{r!(n-r)!} $$ To simplify this, we perform "algebraic surgery" on $r^2$ by rewriting it as $r^2 = r(r-1) + r$. This specific form is chosen because $r(r-1)$ is a "falling factorial" that cancels the $r$ and $(r-1)$ terms in the denominator of the combination formula. This transformation splits our sum into two manageable parts: $$ S = \sum_{r=0}^{20} r(r-1) {}^{20}C_r + \sum_{r=0}^{20} r {}^{20}C_r $$ ### The Power of Absorption We now invoke the **absorption identity**, which states that $r \cdot {}^{n}C_r = n \cdot {}^{n-1}C_{r-1}$. Applying this logic twice yields: $$ r(r-1) \cdot {}^{n}C_r = n(n-1) \cdot {}^{n-2}C_{r-2} $$ By applying these identities, we "absorb" the $r$ terms into the binomial coefficients, effectively lowering the index $n$. Note that the summation limits shift because terms where $r < 2$ vanish: $$ \sum_{r=2}^{20} r(r-1) {}^{20}C_r = 20 \cdot 19 \sum_{r=2}^{20} {}^{18}C_{r-2} = 380 \cdot 2^{18} $$ $$ \sum_{r=1}^{20} r {}^{20}C_r = 20 \sum_{r=1}^{20} {}^{19}C_{r-1} = 20 \cdot 2^{19} $$ ### The Final Synthesis We now combine these results using the property $\sum_{k=0}^{n} {}^{n}C_k = 2^n$. Our total sum $S$ is: $$ S = 380 \cdot 2^{18} + 20 \cdot 2^{19} $$ To simplify, we express $2^{19}$ as $2 \cdot 2^{18}$: $$ S = 380 \cdot 2^{18} + 20 \cdot (2 \cdot 2^{18}) $$ $$ S = (380 + 40) \cdot 2^{18} $$ The final result is: $$ \mathbf{S = 420 \cdot 2^{18}} $$

JEE Main 2021 (26 Aug Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: 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 :

Q: Why can't we just use the binomial expansion of $(1+x)^{20}$ and differentiate it twice?

You absolutely can! Differentiating $(1+x)^{20}$ twice with respect to $x$ and then setting $x=1$ is a perfectly valid and elegant method. However, the algebraic method of 'falling factorials' (using $r(r-1)$) is often preferred in competitive exams because it avoids the potential pitfalls of evaluating derivatives at specific points and is more robust when dealing with more complex summations.

Q: Why do the summation limits change from $r=0$ to $r=2$ in the first part?

Great observation! In the term $r(r-1) {}^{20}C_r$, when $r=0$, the term is $0 \cdot (-1) \cdot {}^{20}C_0 = 0$. When $r=1$, the term is $1 \cdot 0 \cdot {}^{20}C_1 = 0$. Since these terms contribute nothing to the sum, we can safely start the summation from $r=2$ without changing the total value. This is a crucial step to ensure the binomial identity ${}^{n-2}C_{r-2}$ is well-defined.

Q: What if the term was $r^3$ instead of $r^2$?

If you had $r^3$, you would simply extend the 'falling factorial' technique. You would write $r^3$ as $r(r-1)(r-2) + 3r(r-1) + r$. This allows you to break the sum into three parts, each of which can be solved using the same absorption identity. It is a powerful, systematic approach for any polynomial in $r$!

Select Answer:

Visualized Solution

Understanding the Goal

Given: $^{20} C_{r}$ is the coefficient of $x^{r}$ in $(1 + x)^{20}$
Goal: Evaluate $S = \sum_{r = 0}^{20} r^{2}^{20} C_{r}$

The Algebraic Trick for $r^{2}$

We need to eliminate $r^{2}$ to use standard binomial sum formulas.
Rewrite $r^{2}$ as: $r^{2} = r (r - 1) + r$

Splitting the Summation

Substitute $r^{2}$ into the summation:
$S = \sum_{r = 0}^{20} (r (r - 1) + r)^{20} C_{r}$
Split into two separate series:
$S = \sum_{r = 0}^{20} r (r - 1)^{20} C_{r} + \sum_{r = 0}^{20} r^{20} C_{r}$

Adjusting the Limits

For $r = 0$ , $r = 0$ and $r (r - 1) = 0$ .
For $r = 1$ , $r (r - 1) = 0$ .
Adjust lower limits to avoid zero terms:
$S = \sum_{r = 2}^{20} r (r - 1)^{20} C_{r} + \sum_{r = 1}^{20} r^{20} C_{r}$

Property of Binomial Coefficients

Key Property: $r \cdot^{n} C_{r} = n \cdot^{n - 1} C_{r - 1}$
Iterative Property: $r (r - 1) \cdot^{n} C_{r} = n (n - 1) \cdot^{n - 2} C_{r - 2}$

Simplifying the Second Sum

Second sum: $\sum_{r = 1}^{20} r^{20} C_{r}$
Apply property: $\sum_{r = 1}^{20} 20 \cdot^{19} C_{r - 1}$
Factor out constant: $20 \sum_{r = 1}^{20}^{19} C_{r - 1}$

Simplifying the First Sum

First sum: $\sum_{r = 2}^{20} r (r - 1)^{20} C_{r}$
Apply property twice: $\sum_{r = 2}^{20} 20 \cdot 19 \cdot^{18} C_{r - 2}$
Factor out constant: $380 \sum_{r = 2}^{20}^{18} C_{r - 2}$

Evaluating the Binomial Sums

Recall standard result: $\sum_{k = 0}^{n}^{n} C_{k} = 2^{n}$
First sum evaluates to: $\sum_{r = 2}^{20}^{18} C_{r - 2} = 2^{18}$
Second sum evaluates to: $\sum_{r = 1}^{20}^{19} C_{r - 1} = 2^{19}$

Combining the Results

Substitute the evaluated sums back into $S$ :
$S = 380 \cdot 2^{18} + 20 \cdot 2^{19}$

Final Calculation

Simplify the powers of $2$ :
$S = 380 \cdot 2^{18} + 20 \cdot (2 \cdot 2^{18})$
$S = 380 \cdot 2^{18} + 40 \cdot 2^{18}$
$S = (380 + 40) \cdot 2^{18}$
$S = 420 \cdot 2^{18}$

Conclusion

Final Answer: $420 \times 2^{18}$ (Matches Option 4)
Key Takeaway: Always express $r^{k}$ in terms of falling factorials like $r (r - 1)$ to absorb $r$ into binomial coefficients.

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The problem asks us to evaluate the sum

S = \sum_{r = 0}^{20} r^{2}^{20} C_{r}

The presence of the

r^{2}

term prevents the direct application of the standard binomial identity

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

. We must transform the expression to isolate the binomial coefficients.

The Algebraic Surgery

The core difficulty is the

r^{2}

term. We recall that the binomial coefficient is defined as:

^{n} C_{r} = \frac{n !}{r ! ( n - r )!}

To simplify this, we perform "algebraic surgery" on

r^{2}

by rewriting it as

r^{2} = r (r - 1) + r

. This specific form is chosen because

r (r - 1)

is a "falling factorial" that cancels the

r

and

(r - 1)

terms in the denominator of the combination formula.

This transformation splits our sum into two manageable parts:

S = r = 0 \sum 20 r (r - 1)^{20} C_{r} + r = 0 \sum 20 r^{20} C_{r}

The Power of Absorption

We now invoke the absorption identity, which states that

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

. Applying this logic twice yields:

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

By applying these identities, we "absorb" the

r

terms into the binomial coefficients, effectively lowering the index

n

. Note that the summation limits shift because terms where

r < 2

vanish:

r = 2 \sum 20 r (r - 1)^{20} C_{r} = 20 \cdot 19 r = 2 \sum 20^{18} C_{r - 2} = 380 \cdot 2^{18}

r = 1 \sum 20 r^{20} C_{r} = 20 r = 1 \sum 20^{19} C_{r - 1} = 20 \cdot 2^{19}

The Final Synthesis

We now combine these results using the property

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

. Our total sum

S

is:

S = 380 \cdot 2^{18} + 20 \cdot 2^{19}

To simplify, we express

2^{19}

2 \cdot 2^{18}

S = 380 \cdot 2^{18} + 20 \cdot (2 \cdot 2^{18})

S = (380 + 40) \cdot 2^{18}

The final result is:

S = 420 \cdot 2^{18}

Common Traps & FAQs

Why can't we just use the binomial expansion of

(1 + x)^{20}

and differentiate it twice?

You absolutely can! Differentiating

(1 + x)^{20}

twice with respect to

x

and then setting

x = 1

is a perfectly valid and elegant method. However, the algebraic method of 'falling factorials' (using

r (r - 1)

) is often preferred in competitive exams because it avoids the potential pitfalls of evaluating derivatives at specific points and is more robust when dealing with more complex summations.

Why do the summation limits change from

r = 0

r = 2

in the first part?

Great observation! In the term

r (r - 1)^{20} C_{r}

, when

r = 0

, the term is

0 \cdot (- 1) \cdot^{20} C_{0} = 0

. When

r = 1

, the term is

1 \cdot 0 \cdot^{20} C_{1} = 0

. Since these terms contribute nothing to the sum, we can safely start the summation from

r = 2

without changing the total value. This is a crucial step to ensure the binomial identity

^{n - 2} C_{r - 2}

is well-defined.

What if the term was

r^{3}

instead of

r^{2}

If you had

r^{3}

, you would simply extend the 'falling factorial' technique. You would write

r^{3}

r (r - 1) (r - 2) + 3 r (r - 1) + r

. This allows you to break the sum into three parts, each of which can be solved using the same absorption identity. It is a powerful, systematic approach for any polynomial in

r

Animated Solution for Mathematics - Binomial Theorem: 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 :

Select Answer:

Visualized Solution

Understanding the Goal

The Algebraic Trick for $r^{2}$

Splitting the Summation

Adjusting the Limits

Property of Binomial Coefficients

Simplifying the Second Sum

Simplifying the First Sum

Evaluating the Binomial Sums

Combining the Results

Final Calculation

Conclusion

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Algebraic Surgery

The Power of Absorption

The Final Synthesis

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 ___

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

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

The value of $\sum_{r=0}^{20} ^{50-r}C_6$ 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

Let $α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1)^{n} C_{r}$ and $β = (\sum_{r = 0}^{n} \frac{^{n} C _{r}}{r + 1}) + \frac{1}{n + 1}$ . If $140 < \frac{2 α}{β} < 281$ , then the value of $n$ is _______

The value of $^{50}C_4 + \sum_{r=1}^6 ^{56-r}C_3$ is

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 20Cr​ is the co-efficient of xr in the expansion of (1+x)20, then the value of ∑r=020​r220Cr​ is equal to :

Select Answer:

Visualized Solution

Understanding the Goal

The Algebraic Trick for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2

Splitting the Summation

Adjusting the Limits

Property of Binomial Coefficients

Simplifying the Second Sum

Simplifying the First Sum

Evaluating the Binomial Sums

Combining the Results

Final Calculation

Conclusion

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Algebraic Surgery

The Power of Absorption

The Final Synthesis

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; } ∑k=020​(20Ck​)2 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 sum of the series 2.20C0​+5.20C1​+8.20C2​+11.20C3​+...+62.20C20​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of \sum_{r=0}^{20} ^{50-r}C_6 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; } Let α=∑r=0n​(4r2+2r+1)nCr​ and β=(∑r=0n​r+1nCr​​)+n+11​. If 140<β2α​<281, then the value of n is _______

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ^{50}C_4 + \sum_{r=1}^6 ^{56-r}C_3 is

Animated Solution for Mathematics - Binomial Theorem: 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 :

The Algebraic Trick for $r^{2}$

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 ___

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

The value of $\sum_{r=0}^{20} ^{50-r}C_6$ 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

Let $α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1)^{n} C_{r}$ and $β = (\sum_{r = 0}^{n} \frac{^{n} C _{r}}{r + 1}) + \frac{1}{n + 1}$ . If $140 < \frac{2 α}{β} < 281$ , then the value of $n$ is _______

The value of $^{50}C_4 + \sum_{r=1}^6 ^{56-r}C_3$ is