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Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^{48}$ in $(1 + x) + 2 (1 + x)^{2} + 3 (1 + x)^{3} + \dots + 100 (1 + x)^{100}$ is equal to

Q: How do I know which power to look for in the numerator when I have denominators like $x$ and $x^2$?

Think of it as a shift. If you have a term like $\frac{1}{x} \cdot (1+x)^{101}$, you are essentially looking for the coefficient of $x^{49}$ in $(1+x)^{101}$ because the $x$ in the denominator will divide that $x^{49}$ down to $x^{48}$. Similarly, for a term with $x^2$ in the denominator, you need an $x^{50}$ in the numerator to result in $x^{48}$. It is just balancing the powers!

Select Answer:

Progressive Hints

Hint 1

The expression

S = (1 + x) + 2 (1 + x)^{2} + 3 (1 + x)^{3} + \dots + 100 (1 + x)^{100}

is not just a random sum; it is an Arithmetico-Geometric Progression (AGP). The coefficients

1, 2, 3, \dots, 100

form an arithmetic progression, while the terms

(1 + x), (1 + x)^{2}, \dots

form a geometric progression. Do not try to expand each term individually; that is a trap.

Hint 2

To simplify this, use the standard AGP summation technique. Let

r = 1 + x

. The series becomes

S = r + 2 r^{2} + 3 r^{3} + \dots + 100 r^{100}

. Now, multiply the entire series by the common ratio

r

and subtract the new series from the original one to collapse the middle terms into a simple geometric series.

Hint 3

After you find the closed-form expression for

S

in terms of

x

, you will have denominators like

x

and

x^{2}

. To find the coefficient of

x^{48}

in the final expansion, you must look for the coefficient of

x^{49}

in the numerator of the first term and the coefficient of

x^{50}

in the numerator of the second term, adjusting for the denominators.

Visualized Solution

Analyzing the Series

The given series is $S = (1 + x) + 2 (1 + x)^{2} + 3 (1 + x)^{3} + \dots + 100 (1 + x)^{100}$ .
We need to find the coefficient of $x^{48}$ in this expansion.
Notice the pattern: coefficients are in AP ( $1, 2, 3 \dots$ ) and terms are in GP ( $(1 + x), (1 + x)^{2} \dots$ ).
This is an Arithmetico-Geometric Progression (AGP).

Simplifying with Substitution

To make the AGP easier to handle, let's substitute $1 + x = r$ .
The series transforms into: $S = 1 r + 2 r^{2} + 3 r^{3} + \dots + 100 r^{100}$ .
Now the AP part is clearly $1, 2, 3, \dots, 100$ .
The GP part is $r, r^{2}, r^{3}, \dots, r^{100}$ with common ratio $r$ .

The AGP Summation Technique

The standard method to sum an AGP is to multiply the series by the common ratio of the GP.
Here, the common ratio is $r$ .
Multiply the entire equation by $r$ :
$r S = 1 r^{2} + 2 r^{3} + \dots + 99 r^{100} + 100 r^{101}$ .
We shift the terms by one position to align like powers of $r$ .

Subtracting the Equations

Subtract the new equation from the original one: $(S - r S)$ .
$S (1 - r) = r + (2 r^{2} - 1 r^{2}) + (3 r^{3} - 2 r^{3}) + \dots - 100 r^{101}$ .
Simplifying the differences:
$S (1 - r) = (r + r^{2} + r^{3} + \dots + r^{100}) - 100 r^{101}$ .

Sum of the Geometric Progression

The bracketed part $(r + r^{2} + \dots + r^{100})$ is a GP.
First term $a = r$ , common ratio $= r$ , number of terms $n = 100$ .
Sum of GP $= \frac{a ( r ^{n} - 1 )}{r - 1} = \frac{r ( r ^{100} - 1 )}{r - 1}$ .
Substituting this back:
$S (1 - r) = \frac{r ( r ^{100} - 1 )}{r - 1} - 100 r^{101}$ .

Reverting to $x$

We need the answer in terms of $x$ , so we substitute $r = 1 + x$ back.
This implies $1 - r = - x$ and $r - 1 = x$ .
Let's carefully substitute these into our equation.
$S (- x) = \frac{( 1 + x ) (( 1 + x ) ^{100} - 1 )}{x} - 100 (1 + x)^{101}$ .

Isolating the Sum $S$

Divide the entire equation by $- x$ to isolate $S$ .
$S = \frac{100 ( 1 + x ) ^{101}}{x} - \frac{( 1 + x ) (( 1 + x ) ^{100} - 1 )}{x ^{2}}$ .
Expanding the second term's numerator:
$S = \frac{100 ( 1 + x ) ^{101}}{x} - \frac{( 1 + x ) ^{101} - ( 1 + x )}{x ^{2}}$ .

Extracting the Coefficient of $x^{48}$

We need the coefficient of $x^{48}$ in $S$ .
$S$ has two main terms. Let's analyze them separately.
Term 1: $\frac{100 ( 1 + x ) ^{101}}{x}$ . To get $x^{48}$ overall, we need the coefficient of $x^{49}$ in the numerator $100 (1 + x)^{101}$ .
Term 2: $\frac{( 1 + x ) ^{101} - ( 1 + x )}{x ^{2}}$ . To get $x^{48}$ overall, we need the coefficient of $x^{50}$ in the numerator $(1 + x)^{101}$ .

Applying the Binomial Theorem

By Binomial Theorem, the coefficient of $x^{k}$ in $(1 + x)^{n}$ is $^{n} C_{k}$ .
In Term 1: Coeff of $x^{49}$ in $100 (1 + x)^{101}$ is $100 \cdot^{101} C_{49}$ .
In Term 2: Coeff of $x^{50}$ in $(1 + x)^{101}$ is $^{101} C_{50}$ .
Subtracting the two gives the final coefficient:
Final Answer: $100 \cdot^{101} C_{49} -^{101} C_{50}$ .

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Solution Diagram

Analyzing the Setup

The given series is

S = (1 + x) + 2 (1 + x)^{2} + 3 (1 + x)^{3} + \dots + 100 (1 + x)^{100}

This is an Arithmetico-Geometric Progression (AGP) where the coefficients

1, 2, 3, \dots, 100

form an arithmetic progression and the terms

(1 + x), (1 + x)^{2}, \dots

form a geometric progression.

The Algebraic Dance

To simplify the expression, let

r = 1 + x

. The series becomes:

S = r + 2 r^{2} + 3 r^{3} + \dots + 100 r^{100}

Multiply the entire series by the common ratio

r

r S = r^{2} + 2 r^{3} + \dots + 99 r^{100} + 100 r^{101}

Subtract

r S

from

S

to collapse the middle terms:

S (1 - r) = r + r^{2} + r^{3} + \dots + r^{100} - 100 r^{101}

The Binomial Bridge

The sum of the geometric progression

r + r^{2} + \dots + r^{100}

is given by

\frac{r ( r ^{100} - 1 )}{r - 1}

. Substituting this into our equation:

S (1 - r) = \frac{r ( r ^{100} - 1 )}{r - 1} - 100 r^{101}

Now, substitute

r = 1 + x

back into the equation, noting that

1 - r = - x

S (- x) = \frac{( 1 + x ) (( 1 + x ) ^{100} - 1 )}{x} - 100 (1 + x)^{101}

Dividing by

- x

isolates

S

S = \frac{100 ( 1 + x ) ^{101}}{x} - \frac{( 1 + x ) ^{101} - ( 1 + x )}{x ^{2}}

The Final Extraction

We seek the coefficient of

x^{48}

in the expansion of

S

For the first term,

\frac{100 ( 1 + x ) ^{101}}{x}

, we require the coefficient of

x^{49}

100 (1 + x)^{101}

, which is

100 \cdot^{101} C_{49}

For the second term,

- \frac{( 1 + x ) ^{101}}{x ^{2}}

, we require the coefficient of

x^{50}

- (1 + x)^{101}

, which is

-^{101} C_{50}

. (Note: the term

- (1 + x)

does not contribute to the

x^{48}

coefficient).

Combining these results, the final coefficient is: $100 \cdot^{101} C_{49} -^{101} C_{50}$

Common Traps & FAQs

Why can't I just expand each term using the binomial theorem and sum them up? It seems more direct.

While technically possible, expanding one hundred separate binomial terms would be a computational nightmare and a recipe for errors. The beauty of the AGP method is that it uses the structure of the series to collapse it into a manageable form. In JEE Advanced, recognizing the pattern is half the battle; the other half is choosing the most efficient path.

Why did we substitute

r = 1 + x

instead of working with

(1 + x)

directly?

Great question! It is purely for cognitive clarity. Working with a single variable

r

reduces the visual clutter and helps you focus on the underlying algebraic structure of the AGP. It prevents you from getting lost in the binomial terms while you are still trying to sum the series. Always simplify the notation when you can!

How do I know which power to look for in the numerator when I have denominators like

x

and

x^{2}

Think of it as a shift. If you have a term like

\frac{1}{x} \cdot (1 + x)^{101}

, you are essentially looking for the coefficient of

x^{49}

(1 + x)^{101}

because the

x

in the denominator will divide that

x^{49}

down to

x^{48}

. Similarly, for a term with

x^{2}

in the denominator, you need an

x^{50}

in the numerator to result in

x^{48}

. It is just balancing the powers!

Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^{48}$ in $(1 + x) + 2 (1 + x)^{2} + 3 (1 + x)^{3} + \dots + 100 (1 + x)^{100}$ is equal to

Select Answer:

Visualized Solution

Analyzing the Series

Simplifying with Substitution

The AGP Summation Technique

Subtracting the Equations

Sum of the Geometric Progression

Reverting to $x$

Isolating the Sum $S$

Extracting the Coefficient of $x^{48}$

Applying the Binomial Theorem

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Algebraic Dance

The Binomial Bridge

The Final Extraction

Similar Questions

The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + \dots + x^{1000}$ is :

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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

Let $m$ be the smallest positive integer such that the coefficient of $x^{2}$ in the expansion of $(1 + x)^{2} + (1 + x)^{3} + ... + (1 + x)^{49} + (1 + m x)^{50}$ is $(3 n + 1)^{51} C_{3}$ for some positive integer $n$ . Then the value of $n$ is .........

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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

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Let $(1 + x + x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20}$ . If $(a_{1} + a_{3} + a_{5} + \dots + a_{19}) - 11 a_{2} = 121 k$ , then $k$ 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; } The coefficient of x48 in (1+x)+2(1+x)2+3(1+x)3+⋯+100(1+x)100 is equal to

Select Answer:

Visualized Solution

Analyzing the Series

Simplifying with Substitution

The AGP Summation Technique

Subtracting the Equations

Sum of the Geometric Progression

Reverting to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Isolating the Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Extracting the Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x48

Applying the Binomial Theorem

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Algebraic Dance

The Binomial Bridge

The Final Extraction

Similar Questions

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.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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of coefficients of integral power of x in the binomial expansion (1−2x​)50 is

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Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^{48}$ in $(1 + x) + 2 (1 + x)^{2} + 3 (1 + x)^{3} + \dots + 100 (1 + x)^{100}$ is equal to

Reverting to $x$

Isolating the Sum $S$

Extracting the Coefficient of $x^{48}$

The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + \dots + x^{1000}$ is :

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 sum of the coefficients of three consecutive terms in the binomial expansion of $(1 + x)^{n + 2}$ , which are in the ratio $1 : 3 : 5$ , is equal to

The sum of coefficients of integral power of $x$ in the binomial expansion $(1 - 2 x)^{50}$ is

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

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 ______.

Let $(1 + x + x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20}$ . If $(a_{1} + a_{3} + a_{5} + \dots + a_{19}) - 11 a_{2} = 121 k$ , then $k$ is equal to .