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 : Options: (A) $2^{24}$ (B) $2^{25}$ (C) $2^{26}$ (D) $2^{23}$

### Analyzing the Setup The given series is $2 \cdot {}^{20}C_0 + 5 \cdot {}^{20}C_1 + 8 \cdot {}^{20}C_2 + \dots + 62 \cdot {}^{20}C_{20}$. At first glance, it appears complex, but the coefficients $2, 5, 8, 11, \dots, 62$ form an Arithmetic Progression (A.P.). This A.P. has a first term $a = 2$ and a common difference $d = 3$. The general term of this sequence is given by $a + rd = 2 + 3r$. By multiplying this by the binomial coefficient ${}^{20}C_r$, we define the general term of the series as $T_r = (3r + 2) \cdot {}^{20}C_r$. We can express the total sum $S$ using sigma notation: $$S = \sum_{r=0}^{20} (3r + 2) \cdot {}^{20}C_r$$ ### Breaking Down the Complexity We can distribute the ${}^{20}C_r$ term inside the bracket to rewrite the sum as $S = \sum_{r=0}^{20} (3r \cdot {}^{20}C_r + 2 \cdot {}^{20}C_r)$. By the linearity of summation, we split this into two manageable parts: $$S = 3 \sum_{r=0}^{20} r \cdot {}^{20}C_r + 2 \sum_{r=0}^{20} {}^{20}C_r$$ ### The Toolkit of Identities To solve this, we utilize two fundamental binomial identities where $n = 20$: 1. The sum of all binomial coefficients: $\sum_{r=0}^{n} {}^{n}C_r = 2^n$. 2. The sum of $r$ times the binomial coefficient: $\sum_{r=1}^{n} r \cdot {}^{n}C_r = n \cdot 2^{n-1}$. Applying these to our expression, the second part becomes $2 \cdot 2^{20}$. For the first part, we apply the second identity to get $3 \cdot (20 \cdot 2^{19})$. ### The Final Elegance We are now left with the algebraic expression: $$S = 3(20 \cdot 2^{19}) + 2(2^{20})$$ Simplifying the first term, we have $60 \cdot 2^{19}$. To simplify the second term, we express $2^{20}$ as $2 \cdot 2^{19}$, yielding $2 \cdot (2 \cdot 2^{19}) = 4 \cdot 2^{19}$. Combining these, we get: $$S = 60 \cdot 2^{19} + 4 \cdot 2^{19} = (60 + 4) \cdot 2^{19} = 64 \cdot 2^{19}$$ Since $64 = 2^6$, the final result is $2^6 \cdot 2^{19} = \mathbf{2^{25}}$.

Why does the sum of $r \cdot {}^{n}C_r$ start at $r=0$ or $r=1$? Does it matter?

It does not matter! If you look at the term for $r=0$, it is $0 \cdot {}^{n}C_0$, which is $0$. So, whether you start the summation at $r=0$ or $r=1$, the value of the sum remains identical. It is a common trick to start at $r=1$ to make the algebra cleaner, but starting at $r=0$ is perfectly valid.

JEE Main 2019 (8 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: 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 :

Q: How do I handle the powers of $2$ when adding $60 \cdot 2^{19} + 4 \cdot 2^{19}$?

Think of $2^{19}$ as a common variable, like $x$. The expression is $60x + 4x$, which is $64x$. Since $64$ is $2^6$, you get $2^6 \cdot 2^{19}$. Using the laws of exponents, you simply add the powers: $6 + 19 = 25$. It is all about finding that common factor!

Select Answer:

Visualized Solution

Observe the Series Structure

Given series: $2 \cdot^{20} C_{0} + 5 \cdot^{20} C_{1} + 8 \cdot^{20} C_{2} + \dots + 62 \cdot^{20} C_{20}$
Notice each term is a product of an integer and a binomial coefficient.

Identify the Arithmetic Progression

The coefficients are $2, 5, 8, 11, \dots, 62$
This sequence forms an Arithmetic Progression (A.P.)
First term ( $a$ ) = $2$
Common difference ( $d$ ) = $5 - 2 = 3$

Define the General Term $T_{r}$

General term of the A.P.: $a + r d = 2 + 3 r$
General term of the series: $T_{r} = (3 r + 2) \cdot^{20} C_{r}$
Total sum $S = \sum_{r = 0}^{20} (3 r + 2) \cdot^{20} C_{r}$

Split the Summation

Expand the general term: $S = \sum_{r = 0}^{20} (3 r \cdot^{20} C_{r} + 2 \cdot^{20} C_{r})$
Split into two sums: $S = 3 \sum_{r = 0}^{20} r \cdot^{20} C_{r} + 2 \sum_{r = 0}^{20}^{20} C_{r}$

Recall Standard Binomial Identities

Identity 1: $\sum_{r = 0}^{n}^{n} C_{r} = 2^{n}$
Identity 2: $\sum_{r = 1}^{n} r \cdot^{n} C_{r} = n \cdot 2^{n - 1}$
Note: For $r = 0$ , $r \cdot^{n} C_{r} = 0$ , so $\sum_{r = 0}^{n} r \cdot^{n} C_{r} = \sum_{r = 1}^{n} r \cdot^{n} C_{r}$

Apply Identities to the Problem

For $n = 20$ :
$\sum_{r = 0}^{20}^{20} C_{r} = 2^{20}$
$\sum_{r = 0}^{20} r \cdot^{20} C_{r} = 20 \cdot 2^{19}$

Substitute Results into the Sum Equation

Recall: $S = 3 \sum_{r = 0}^{20} r \cdot^{20} C_{r} + 2 \sum_{r = 0}^{20}^{20} C_{r}$
Substitute the evaluated sums:
$S = 3 (20 \cdot 2^{19}) + 2 (2^{20})$

Simplify the First Term

Focus on the first part: $3 \times (20 \cdot 2^{19})$
Multiply the constants: $3 \times 20 = 60$
First term becomes: $60 \cdot 2^{19}$
Equation updates to: $S = 60 \cdot 2^{19} + 2 \cdot 2^{20}$

Adjust Powers of $2$ for Factoring

We have $S = 60 \cdot 2^{19} + 2 \cdot 2^{20}$
To add these, make the powers of $2$ identical.
Rewrite $2^{20}$ as $2 \cdot 2^{19}$
Second term becomes: $2 \cdot (2 \cdot 2^{19}) = 4 \cdot 2^{19}$
$S = 60 \cdot 2^{19} + 4 \cdot 2^{19}$

Combine and Conclude

Factor out $2^{19}$ : $S = (60 + 4) \cdot 2^{19}$
$S = 64 \cdot 2^{19}$
Express $64$ as a power of $2$ : $64 = 2^{6}$
$S = 2^{6} \cdot 2^{19} = 2^{25}$
Final Answer: $2^{25}$

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The given series is

2 \cdot^{20} C_{0} + 5 \cdot^{20} C_{1} + 8 \cdot^{20} C_{2} + \dots + 62 \cdot^{20} C_{20}

. At first glance, it appears complex, but the coefficients

2, 5, 8, 11, \dots, 62

form an Arithmetic Progression (A.P.).

This A.P. has a first term

a = 2

and a common difference

d = 3

. The general term of this sequence is given by

a + r d = 2 + 3 r

By multiplying this by the binomial coefficient

^{20} C_{r}

, we define the general term of the series as

T_{r} = (3 r + 2) \cdot^{20} C_{r}

. We can express the total sum

S

using sigma notation:

S = r = 0 \sum 20 (3 r + 2) \cdot^{20} C_{r}

Breaking Down the Complexity

We can distribute the

^{20} C_{r}

term inside the bracket to rewrite the sum as

S = \sum_{r = 0}^{20} (3 r \cdot^{20} C_{r} + 2 \cdot^{20} C_{r})

By the linearity of summation, we split this into two manageable parts:

S = 3 r = 0 \sum 20 r \cdot^{20} C_{r} + 2 r = 0 \sum 20^{20} C_{r}

The Toolkit of Identities

To solve this, we utilize two fundamental binomial identities where

n = 20

1. The sum of all binomial coefficients:

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

. 2. The sum of

r

times the binomial coefficient:

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

Applying these to our expression, the second part becomes

2 \cdot 2^{20}

. For the first part, we apply the second identity to get

3 \cdot (20 \cdot 2^{19})

The Final Elegance

We are now left with the algebraic expression:

S = 3 (20 \cdot 2^{19}) + 2 (2^{20})

Simplifying the first term, we have

60 \cdot 2^{19}

. To simplify the second term, we express

2^{20}

2 \cdot 2^{19}

, yielding

2 \cdot (2 \cdot 2^{19}) = 4 \cdot 2^{19}

Combining these, we get:

S = 60 \cdot 2^{19} + 4 \cdot 2^{19} = (60 + 4) \cdot 2^{19} = 64 \cdot 2^{19}

Since

64 = 2^{6}

, the final result is

2^{6} \cdot 2^{19} = 2^{25}

Common Traps & FAQs

Wait, why do we reject the negative root here? Or rather, why can't we just use the sum of an AP formula?

That is a great question! The reason we cannot use the standard sum of an AP formula (like

\frac{n}{2} (a + l)

) is that the terms are not just an AP; they are an AP *multiplied* by binomial coefficients. The binomial coefficients

^{20} C_{r}

change with each term, so the standard AP sum formula does not apply. We must use the properties of binomial series instead.

Why does the sum of

r \cdot^{n} C_{r}

start at

r = 0

r = 1

? Does it matter?

It does not matter! If you look at the term for

r = 0

, it is

0 \cdot^{n} C_{0}

, which is

0

. So, whether you start the summation at

r = 0

r = 1

, the value of the sum remains identical. It is a common trick to start at

r = 1

to make the algebra cleaner, but starting at

r = 0

is perfectly valid.

How do I handle the powers of

2

when adding

60 \cdot 2^{19} + 4 \cdot 2^{19}

Think of

2^{19}

as a common variable, like

x

. The expression is

60 x + 4 x

, which is

64 x

. Since

64

2^{6}

, you get

2^{6} \cdot 2^{19}

. Using the laws of exponents, you simply add the powers:

6 + 19 = 25

. It is all about finding that common factor!

Animated Solution for Mathematics - Binomial Theorem: 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 :

Select Answer:

Visualized Solution

Observe the Series Structure

Identify the Arithmetic Progression

Define the General Term $T_{r}$

Split the Summation

Recall Standard Binomial Identities

Apply Identities to the Problem

Substitute Results into the Sum Equation

Simplify the First Term

Adjust Powers of $2$ for Factoring

Combine and Conclude

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Breaking Down the Complexity

The Toolkit of Identities

The Final Elegance

Similar Questions

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If $n \geq 2$ is a positive integer, then the sum of the series $^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + ... +^{n} C_{2})$ is :

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

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})$ is:

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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; } The sum of the series 2.20C0​+5.20C1​+8.20C2​+11.20C3​+...+62.20C20​ is equal to :

Select Answer:

Visualized Solution

Observe the Series Structure

Identify the Arithmetic Progression

Define the General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr​

Split the Summation

Recall Standard Binomial Identities

Apply Identities to the Problem

Substitute Results into the Sum Equation

Simplify the First Term

Adjust Powers of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2 for Factoring

Combine and Conclude

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Breaking Down the Complexity

The Toolkit of Identities

The Final Elegance

Similar Questions

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Animated Solution for Mathematics - Binomial Theorem: 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 :

Define the General Term $T_{r}$

Adjust Powers of $2$ for Factoring

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