JEE Advanced 1983

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Animated Solution for Mathematics - Binomial Theorem: If $(1 + a x)^{n} = 1 + 8 x + 24 x^{2} + ....$ then $a = .........$ and $n = .........$

Q: Wait, why do we substitute $ax$ for $x$ in the binomial theorem?

The binomial theorem $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$ is a general template. When you have $(1 + ax)^n$, the term $ax$ acts as the variable. By replacing every instance of $x$ in the template with $ax$, you correctly map the specific problem to the general formula.

Visualized Solution

Identifying the Expansion of $(1 + a x)^{n}$

Given Equation: $(1 + a x)^{n} = 1 + 8 x + 24 x^{2} + \dots$
Objective: Find the values of $a$ and $n$ .

Recalling the Binomial Theorem for $(1 + x)^{n}$

Standard Formula:
$(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots$

Applying the Theorem to $(1 + a x)^{n}$

Substituting $x \to a x$ :
$(1 + a x)^{n} = 1 + n (a x) + \frac{n ( n - 1 )}{2 !} (a x)^{2} + \dots$
Simplified Expansion:
$(1 + a x)^{n} = 1 + na x + \frac{n ( n - 1 ) a ^{2}}{2} x^{2} + \dots$

Comparing the Coefficients of $x$

Comparing with: $1 + 8 x + 24 x^{2} + \dots$
Coefficient of $x$ : $na = 8$

Comparing the Coefficients of $x^{2}$

Coefficient of $x^{2}$ : $\frac{n ( n - 1 ) a ^{2}}{2} = 24$
Simplifying: $n (n - 1) a^{2} = 48$

Expressing $a$ in terms of $n$

From Equation 1: $na = 8$
Isolating $a$ : $a = \frac{8}{n}$

Substituting $a$ into Equation 2

Equation 2: $n (n - 1) a^{2} = 48$
Substituting $a = \frac{8}{n}$ :
$n (n - 1) (\frac{8}{n})^{2} = 48$

Simplifying the Algebraic Equation

Expanding the square: $n (n - 1) \frac{64}{n ^{2}} = 48$
Cancelling $n$ : $\frac{64 ( n - 1 )}{n} = 48$

Solving for $n$

Dividing by 16: $\frac{4 ( n - 1 )}{n} = 3$
Cross-multiplying: $4 (n - 1) = 3 n$
Expanding: $4 n - 4 = 3 n$
Result: $n = 4$

Solving for $a$

Recall: $a = \frac{8}{n}$
Substitute $n = 4$ : $a = \frac{8}{4}$
Result: $a = 2$

Final Conclusion for $a$ and $n$

Final Answer: $a = 2, n = 4$
Verification: $(1 + 2 x)^{4} = 1 + 4 (2 x) + \frac{4 \times 3}{2} (2 x)^{2} + \dots$
$= 1 + 8 x + 24 x^{2} + \dots$

00:00 / 00:00

The Sigma Insight: Binomial Expansion for Positive Integral Index

The Binomial Foundation

The Binomial Theorem serves as a bridge between finite expressions and infinite series. Consider the expression

(1 + a x)^{n}

. While it appears simple, it contains a structured expansion governed by the general theorem:

(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots

By substituting

a x

for

x

, we adapt this master key to our specific problem:

(1 + a x)^{n} = 1 + na x + \frac{n ( n - 1 ) a ^{2}}{2} x^{2} + \dots

Analyzing the Coefficients

We are given the expansion

1 + 8 x + 24 x^{2} + \dots

. By comparing the coefficients of this expression with our expanded form, we establish a system of equations:

na = 8

\frac{n ( n - 1 ) a ^{2}}{2} = 24

Solving the System

To solve for the variables, we first isolate

a

from the linear equation:

a = \frac{8}{n}

Next, we substitute this expression for

a

into the second equation:

\frac{n ( n - 1 )}{2} \cdot (\frac{8}{n})^{2} = 24

This simplifies to:

\frac{n ( n - 1 )}{2} \cdot \frac{64}{n ^{2}} = 24

\frac{32 ( n - 1 )}{n} = 24

Final Calculation

Solving the simplified equation

32 (n - 1) = 24 n

leads us to:

32 n - 32 = 24 n

8 n = 32 \Rightarrow n = 4

Substituting

n = 4

back into our expression for

a

, we find:

a = \frac{8}{4} = 2

The values that satisfy the given expansion are $n = 4$ and $a = 2$ .

Common Traps & FAQs

Wait, why do we substitute

a x

for

x

in the binomial theorem?

The binomial theorem

(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots

is a general template. When you have

(1 + a x)^{n}

, the term

a x

acts as the variable. By replacing every instance of

x

in the template with

a x

, you correctly map the specific problem to the general formula.

Why can't we just use the general term formula

T_{r + 1} =^{n} C_{r} x^{r}

You absolutely can! The general term formula is just another way to write the coefficients. Using

T_{2} =^{n} C_{1} (a x)^{1} = na x

and

T_{3} =^{n} C_{2} (a x)^{2} = \frac{n ( n - 1 )}{2} a^{2} x^{2}

leads to the exact same equations. It is a matter of preference, but both paths lead to the same destination.

What if the expansion had more terms, like

x^{3}

That is a great question! If the problem provided the coefficient for

x^{3}

, you would simply have a third equation. This would allow you to solve for three variables if needed, or verify your values of

a

and

n

to ensure they are consistent across all terms.

Animated Solution for Mathematics - Binomial Theorem: If $(1 + a x)^{n} = 1 + 8 x + 24 x^{2} + ....$ then $a = .........$ and $n = .........$

Visualized Solution

Identifying the Expansion of $(1 + a x)^{n}$

Recalling the Binomial Theorem for $(1 + x)^{n}$

Applying the Theorem to $(1 + a x)^{n}$

Comparing the Coefficients of $x$

Comparing the Coefficients of $x^{2}$

Expressing $a$ in terms of $n$

Substituting $a$ into Equation 2

Simplifying the Algebraic Equation

Solving for $n$

Solving for $a$

Final Conclusion for $a$ and $n$

The Sigma Insight: Binomial Expansion for Positive Integral Index

The Binomial Foundation

Analyzing the Coefficients

Solving the System

Final Calculation

Similar Questions

If in the expansion of $(1 + x)^{m} (1 - x)^{n}$ , the coefficients of $x$ and $x^{2}$ are 3 and $- 6$ respectively, then $m$ is

The coefficient of $x^{n}$ in expansion of $(1 + x) (1 - x)^{n}$ is

Let the sum of the coefficients of the first three terms in the expansion of $(x - \frac{3}{x ^{2}})^{n}, x \neq = 0, n \in N$ , be 376. Then the coefficient of $x^{4}$ is ______.

If the coefficients of $x$ and $x^{2}$ in $(1 + x)^{p} (1 - x)^{q}$ are 4 and $- 5$ respectively, then $2 p + 3 q$ is equal to

If n is the degree of the polynomial, $[\frac{2}{5 x ^{3} + 1 - 5 x ^{3} - 1}]^{8} + [\frac{2}{5 x ^{3} + 1 + 5 x ^{3} - 1}]^{8}$ and m is the coefficient of $x^{n}$ in it, then the ordered pair (n,m) is equal to:

The coefficient of $x^{2012}$ in the expansion of $(1 - x)^{2008} (1 + x + x^{2})^{2007}$ is equal to

For natural numbers $m, n$ if $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + ....$ and $a_{1} = a_{2} = 10$ , then $(m, n)$ is

If in the expansion of $(1 + x)^{p} (1 - x)^{q}$ the coefficients of $x$ and $x^{2}$ are 1 and -2, respectively, then $p^{2} + q^{2}$ is equal to:

If the coefficient of $x$ in the expansion of $(a x^{2} + b x + c) (1 - 2 x)^{26}$ is -56 and the coefficients of $x^{2}$ and $x^{3}$ are both zero, then $a + b + c$ is equal to :

The coefficient of $x^{2}$ in the expansion of the product $(2 - x^{2}) \cdot ((1 + 2 x + 3 x^{2})^{6} + (1 - 4 x^{2})^{6})$ 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 (1+ax)n=1+8x+24x2+.... then a=......... and n=.........

Visualized Solution

Identifying the Expansion of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1+ax)n

Recalling the Binomial Theorem for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1+x)n

Applying the Theorem to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1+ax)n

Comparing the Coefficients of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Comparing the Coefficients of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a into Equation 2

Simplifying the Algebraic Equation

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

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

Final Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

The Sigma Insight: Binomial Expansion for Positive Integral Index

The Binomial Foundation

Analyzing the Coefficients

Solving the System

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If in the expansion of (1+x)m(1−x)n, the coefficients of x and x2 are 3 and −6 respectively, then m is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of xn in expansion of (1+x)(1−x)n is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the sum of the coefficients of the first three terms in the expansion of (x−x23​)n,x=0,n∈N, be 376. Then the coefficient of x4 is ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the coefficients of x and x2 in (1+x)p(1−x)q are 4 and −5 respectively, then 2p+3q is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If n is the degree of the polynomial, [5x3+1​−5x3−1​2​]8+[5x3+1​+5x3−1​2​]8 and m is the coefficient of xn in it, then the ordered pair (n,m) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x2012 in the expansion of (1−x)2008(1+x+x2)2007 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For natural numbers m,n if (1−y)m(1+y)n=1+a1​y+a2​y2+.... and a1​=a2​=10, then (m,n) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If in the expansion of (1+x)p(1−x)q the coefficients of x and x2 are 1 and -2, respectively, then p2+q2 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the coefficient of x in the expansion of (ax2+bx+c)(1−2x)26 is -56 and the coefficients of x2 and x3 are both zero, then a+b+c is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x2 in the expansion of the product (2−x2)⋅((1+2x+3x2)6+(1−4x2)6) is :

Animated Solution for Mathematics - Binomial Theorem: If $(1 + a x)^{n} = 1 + 8 x + 24 x^{2} + ....$ then $a = .........$ and $n = .........$

Identifying the Expansion of $(1 + a x)^{n}$

Recalling the Binomial Theorem for $(1 + x)^{n}$

Applying the Theorem to $(1 + a x)^{n}$

Comparing the Coefficients of $x$

Comparing the Coefficients of $x^{2}$

Expressing $a$ in terms of $n$

Substituting $a$ into Equation 2

Solving for $n$

Solving for $a$

Final Conclusion for $a$ and $n$

If in the expansion of $(1 + x)^{m} (1 - x)^{n}$ , the coefficients of $x$ and $x^{2}$ are 3 and $- 6$ respectively, then $m$ is

The coefficient of $x^{n}$ in expansion of $(1 + x) (1 - x)^{n}$ is

Let the sum of the coefficients of the first three terms in the expansion of $(x - \frac{3}{x ^{2}})^{n}, x \neq = 0, n \in N$ , be 376. Then the coefficient of $x^{4}$ is ______.

If the coefficients of $x$ and $x^{2}$ in $(1 + x)^{p} (1 - x)^{q}$ are 4 and $- 5$ respectively, then $2 p + 3 q$ is equal to

If n is the degree of the polynomial, $[\frac{2}{5 x ^{3} + 1 - 5 x ^{3} - 1}]^{8} + [\frac{2}{5 x ^{3} + 1 + 5 x ^{3} - 1}]^{8}$ and m is the coefficient of $x^{n}$ in it, then the ordered pair (n,m) is equal to:

The coefficient of $x^{2012}$ in the expansion of $(1 - x)^{2008} (1 + x + x^{2})^{2007}$ is equal to

For natural numbers $m, n$ if $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + ....$ and $a_{1} = a_{2} = 10$ , then $(m, n)$ is

If in the expansion of $(1 + x)^{p} (1 - x)^{q}$ the coefficients of $x$ and $x^{2}$ are 1 and -2, respectively, then $p^{2} + q^{2}$ is equal to:

If the coefficient of $x$ in the expansion of $(a x^{2} + b x + c) (1 - 2 x)^{26}$ is -56 and the coefficients of $x^{2}$ and $x^{3}$ are both zero, then $a + b + c$ is equal to :

The coefficient of $x^{2}$ in the expansion of the product $(2 - x^{2}) \cdot ((1 + 2 x + 3 x^{2})^{6} + (1 - 4 x^{2})^{6})$ is :