Find the term independent of $x$, $x \neq 0$, in the expansion of $\left(\dfrac{3x^2}{2} - \dfrac{1}{3x}\right)^{15}$.

> [!TLDR] > To find the term independent of $x$, we use the general term of the binomial expansion, set the power of $x$ to zero to find the term number, and then calculate its coefficient. The term independent of $x$ is $\boxed{\frac{1001}{2592}}$. When expanding a binomial expression like $(a+b)^n$, each term will generally contain powers of $a$ and $b$. If $a$ and $b$ themselves contain variables like $x$, then each term will have a specific power of $x$. Our goal is to find the term where the power of $x$ is zero, meaning $x^0 = 1$, so $x$ effectively disappears from that term. This is what "independent of $x$" means. The key to solving this problem lies in the Binomial Theorem, specifically the formula for the general term of an expansion. > [!FORMULA] > The general term, $T_{r+1}$, in the binomial expansion of $(a+b)^n$ is given by: > $$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$ > where $r$ is an integer from $0$ to $n$. Let's break down the problem step-by-step. 1. **Identify the components of the binomial expression.** The given expression is $\left(\dfrac{3x^2}{2} - \dfrac{1}{3x}\right)^{15}$. Comparing this to $(a+b)^n$: * $a = \dfrac{3x^2}{2}$ * $b = -\dfrac{1}{3x}$ * $n = 15$ 2. **Write down the general term $T_{r+1}$.** Substitute $a$, $b$, and $n$ into the general term formula: $$T_{r+1} = \binom{15}{r} \left(\dfrac{3x^2}{2}\right)^{15-r} \left(-\dfrac{1}{3x}\right)^r$$ Now, we need to separate the numerical coefficients from the powers of $x$. $$T_{r+1} = \binom{15}{r} \left(\dfrac{3}{2}\right)^{15-r} (x^2)^{15-r} \left(-\dfrac{1}{3}\right)^r \left(\dfrac{1}{x}\right)^r$$ Using exponent rules $(x^m)^n = x^{mn}$ and $\dfrac{1}{x^k} = x^{-k}$: $$T_{r+1} = \binom{15}{r} \left(\dfrac{3}{2}\right)^{15-r} x^{2(15-r)} \left(-\dfrac{1}{3}\right)^r x^{-r}$$ 3. **Combine all terms involving $x$.** The powers of $x$ are $x^{2(15-r)}$ and $x^{-r}$. When multiplying terms with the same base, we add their ex

Find the term independent of x, x ≠ 0, in the expa – Class 11 Mathematics

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Concept understanding — Binomial Theorem Expansion

The Problem That Started It All

Imagine you have to expand $(x + y)^2$ . That's easy: $x^2 + 2xy + y^2$ . Now try $(x + y)^3$ : $x^3 + 3x^2y + 3xy^2 + y^3$ . Still manageable.

But what about $(x + y)^{10}$ ? Or $(x + y)^{100}$ ? Multiplying it out term by term would take forever. There has to be a pattern — and there is.

The Binomial Theorem is the shortcut that tells you exactly what each term in the expansion of $(x + y)^n$ looks like, without ever having to multiply.

The Pattern You Already Know

Look at the expansions you already know:

Power	Expansion
$(x + y)^0$	$1$
$(x + y)^1$	$x + y$
$(x + y)^2$	$x^2 + 2xy + y^2$
$(x + y)^3$	$x^3 + 3x^2y + 3xy^2 + y^3$
$(x + y)^4$	$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

Notice three things:

The powers of $x$ decrease from $n$ down to $0$ , while the powers of $y$ increase from $0$ up to $n$ . In every term, the exponents add to $n$ .
The coefficients — 1, 4, 6, 4, 1 for $n=4$ — follow a famous pattern called Pascal's triangle.
The number of terms is always $n+1$ .

Note

Pascal's triangle: each number is the sum of the two numbers directly above it.

       1
      1 1
     1 2 1
    1 3 3 1
   1 4 6 4 1
  1 5 10 10 5 1

The Precise Statement

n

Let's break that down.

The symbol $\sum_{k=0}^{n}$ means "add up terms for $k = 0, 1, 2, \dots, n$ ". For each $k$ , the term is:

$\binom{n}{k}$ — the coefficient (read as "n choose k")
$x^{n-k}$ — $x$ raised to the power $n-k$
$y^k$ — $y$ raised to the power $k$

So for $n=4$ , the terms are:

$k$	$\binom{4}{k}$	$x^{4-k}$	$y^k$	Term
0	$\binom{4}{0}=1$	$x^4$	$y^0$	$1 \cdot x^4$
1	$\binom{4}{1}=4$	$x^3$	$y^1$	$4x^3y$
2	$\binom{4}{2}=6$	$x^2$	$y^2$	$6x^2y^2$
3	$\binom{4}{3}=4$	$x^1$	$y^3$	$4xy^3$
4	$\binom{4}{4}=1$	$x^0$	$y^4$	$1 \cdot y^4$

Add them up: $x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$ . Exactly what we had.

Where Do Those Coefficients Come From?

The binomial coefficient $\binom{n}{k}$ counts how many ways you can choose $k$ items from a set of $n$ items. In the expansion, it counts how many ways you can pick $k$ copies of $y$ (and therefore $n-k$ copies of $x$ ) when multiplying $(x+y)$ by itself $n$ times.

Tip

To compute $\binom{n}{k}$ quickly: start at $n$ and multiply $k$ decreasing numbers, then divide by $k!$ .

Example: $\binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35$ .

The General Term

The $k$ -th term (starting with $k=0$ ) in the expansion is:

\text{General term: } T_{k+1} = \binom{n}{k} x^{n-k} y^k

This is the most useful part for exams. If someone asks "find the 5th term in $(x+y)^{10}$ ", you set $k=4$ (because $T_{k+1}$ means $k=4$ gives the 5th term) and write:

T_5 = \binom{10}{4} x^{6} y^{4}

What If It's Not Just $x$ and $y$ ?

The theorem works for any expression. For $(2a - 3b)^5$ , treat $x = 2a$ and $y = -3b$ :

(2a - 3b)^5 = \sum_{k=0}^{5} \binom{5}{k} (2a)^{5-k} (-3b)^k

The $k$ -th term becomes $\binom{5}{k} (2)^{5-k} (-3)^k a^{5-k} b^k$ . The coefficients get multiplied by powers of 2 and -3.

Watch out

A common mistake: forgetting the sign. If the second term is negative, every odd $k$ (1, 3, 5, ...) picks up a negative sign from $(-3)^k$ .

Why This Matters

The Binomial Theorem isn't just a formula — it's a window into combinatorics, probability, and even calculus. It lets you:

Expand any binomial instantly
Find a specific term without expanding everything
Approximate values like $(1.01)^{10}$ by setting $x=1$ , $y=0.01$
Understand the binomial distribution in statistics

The core idea: every term in $(x+y)^n$ has the form $\binom{n}{k} x^{n-k} y^k$ , and the theorem tells you exactly which $k$ to use and what coefficient goes with it.

Expanding binomial expressions using the Binomial Theorem is a core topic in the NCERT Class 11 Mathematics chapter on Binomial Theorem, and "binomial theorem expansion formula and examples" is one of the most searched topics for CBSE board and JEE Main preparation. Finding a specific general term without full expansion is also a classic question type that appears repeatedly in "binomial theorem important questions" for competitive exams.

The key idea is to use the Binomial Theorem Expansion to find the general term.

The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ .

For $\left(\dfrac{3x^2}{2} - \dfrac{1}{3x}\right)^{15}$ , we have $a = \dfrac{3x^2}{2}$ , $b = -\dfrac{1}{3x}$ , and $n = 15$ .

Write the general term $T_{r+1}$ :

T_{r+1} = \binom{15}{r} \left(\frac{3x^2}{2}\right)^{15-r} \left(-\frac{1}{3x}\right)^r

Simplify the powers of $x$ :

T_{r+1} = \binom{15}{r} \left(\frac{3}{2}\right)^{15-r} (x^2)^{15-r} \left(-\frac{1}{3}\right)^r (x^{-1})^r

T_{r+1} = \binom{15}{r} \left(\frac{3}{2}\right)^{15-r} \left(-\frac{1}{3}\right)^r x^{2(15-r) - r}

T_{r+1} = \binom{15}{r} \left(\frac{3}{2}\right)^{15-r} \left(-\frac{1}{3}\right)^r x^{30 - 2r - r}

T_{r+1} = \binom{15}{r} \left(\frac{3}{2}\right)^{15-r} \left(-\frac{1}{3}\right)^r x^{30 - 3r}

For the term independent of $x$ , the exponent of $x$ must be zero:

30 - 3r = 0 \implies 3r = 30 \implies r = 10

Substitute $r=10$ back into the general term (excluding $x$ ):

T_{10+1} = T_{11} = \binom{15}{10} \left(\frac{3}{2}\right)^{15-10} \left(-\frac{1}{3}\right)^{10}

T_{11} = \binom{15}{10} \left(\frac{3}{2}\right)^5 \left(\frac{1}{3}\right)^{10}

T_{11} = \binom{15}{5} \frac{3^5}{2^5 \cdot 3^{10}} = \binom{15}{5} \frac{1}{2^5 \cdot 3^5}

T_{11} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \times \frac{1}{(2 \times 3)^5}

T_{11} = (3 \times 7 \times 13 \times 11) \times \frac{1}{6^5} = 3003 \times \frac{1}{7776} = \frac{3003}{7776}

Reduce by dividing numerator and denominator by 3: $\dfrac{3003}{7776} = \dfrac{1001}{2592}$.

✓Final answer

x

To find the term independent of $x$ , we use the general term of the binomial expansion, set the power of $x$ to zero to find the term number, and then calculate its coefficient. The term independent of $x$ is $\boxed{\frac{1001}{2592}}$ .

When expanding a binomial expression like $(a+b)^n$ , each term will generally contain powers of $a$ and $b$ . If $a$ and $b$ themselves contain variables like $x$ , then each term will have a specific power of $x$ . Our goal is to find the term where the power of $x$ is zero, meaning $x^0 = 1$ , so $x$ effectively disappears from that term. This is what "independent of $x$ " means.

The key to solving this problem lies in the Binomial Theorem, specifically the formula for the general term of an expansion.

T_{r+1}

Let's break down the problem step-by-step.

Identify the components of the binomial expression.

The given expression is $\left(\dfrac{3x^2}{2} - \dfrac{1}{3x}\right)^{15}$ .

Comparing this to $(a+b)^n$ :
- $a = \dfrac{3x^2}{2}$
- $b = -\dfrac{1}{3x}$
- $n = 15$
Write down the general term $T_{r+1}$ .

Substitute $a$ , $b$ , and $n$ into the general term formula:

T_{r+1} = \binom{15}{r} \left(\dfrac{3x^2}{2}\right)^{15-r} \left(-\dfrac{1}{3x}\right)^r

Now, we need to separate the numerical coefficients from the powers of $x$.

T_{r+1} = \binom{15}{r} \left(\dfrac{3}{2}\right)^{15-r} (x^2)^{15-r} \left(-\dfrac{1}{3}\right)^r \left(\dfrac{1}{x}\right)^r

Using exponent rules $(x^m)^n = x^{mn}$ and $\dfrac{1}{x^k} = x^{-k}$:

T_{r+1} = \binom{15}{r} \left(\dfrac{3}{2}\right)^{15-r} x^{2(15-r)} \left(-\dfrac{1}{3}\right)^r x^{-r}

Combine all terms involving $x$ . The powers of $x$ are $x^{2(15-r)}$ and $x^{-r}$ . When multiplying terms with the same base, we add their exponents: $x^m \cdot x^n = x^{m+n}$ . The combined power of $x$ is $x^{2(15-r) - r}$ . Let's simplify the exponent:

2(15-r) - r = 30 - 2r - r = 30 - 3r

So, the general term can be written as:

T_{r+1} = \binom{15}{r} \left(\dfrac{3}{2}\right)^{15-r} \left(-\dfrac{1}{3}\right)^r x^{30-3r}

Find the value of $r$ for the term independent of $x$ . For a term to be independent of $x$ , its power of $x$ must be zero. Set the exponent of $x$ equal to zero:

30 - 3r = 0

3r = 30

r = 10

Since $r=10$ is an integer between $0$ and $n=15$, it is a valid value for $r$. This means the term independent of $x$ is the $(10+1)^{th}$, or $11^{th}$, term.

5. Calculate the coefficient of the term independent of $x$ .

Substitute $r=10$ back into the numerical part of the general term expression (excluding $x^{30-3r}$ ):

T_{11} = \binom{15}{10} \left(\dfrac{3}{2}\right)^{15-10} \left(-\dfrac{1}{3}\right)^{10}

T_{11} = \binom{15}{10} \left(\dfrac{3}{2}\right)^5 \left(-\dfrac{1}{3}\right)^{10}

Let's calculate each part:
*   $\binom{15}{10} = \binom{15}{15-10} = \binom{15}{5}$

\binom{15}{5} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = \frac{15}{5 \times 3} \times \frac{12}{4 \times 1} \times \frac{14}{2} \times 13 \times 11 = 1 \times 3 \times 7 \times 13 \times 11 = 3003

*   $\left(\dfrac{3}{2}\right)^5 = \dfrac{3^5}{2^5} = \dfrac{243}{32}$
*   $\left(-\dfrac{1}{3}\right)^{10} = \dfrac{(-1)^{10}}{3^{10}} = \dfrac{1}{59049}$ (since the exponent is even, the negative sign becomes positive)

Now, multiply these values:

T_{11} = 3003 \times \dfrac{243}{32} \times \dfrac{1}{59049}

Notice that $243 = 3^5$ and $59049 = 3^{10}$. We can simplify this:

T_{11} = 3003 \times \dfrac{3^5}{32} \times \dfrac{1}{3^{10}}

T_{11} = 3003 \times \dfrac{1}{32 \times 3^{10-5}}

T_{11} = 3003 \times \dfrac{1}{32 \times 3^5}

T_{11} = 3003 \times \dfrac{1}{32 \times 243}

T_{11} = \dfrac{3003}{7776}

Both the numerator and denominator are divisible by 3:

3003 \div 3 = 1001

7776 \div 3 = 2592

So, the simplified fraction is:

T_{11} = \dfrac{1001}{2592}

This fraction cannot be simplified further, as $1001 = 7 \times 11 \times 13$ and $2592 = 2^5 \times 3^4$. They share no common prime factors.

Watch out

A common mistake is to forget the negative sign in $b = -\frac{1}{3x}$ or to incorrectly handle its power. Since the exponent $r=10$ is even, $(-\frac{1}{3})^{10}$ becomes positive. If $r$ were odd, the term would be negative.

✓Final answer

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Q.Find the term independent of xxx, x≠0x \neq 0x=0, in the expansion of (3x22−13x)15\left(\dfrac{3x^2}{2} - \dfrac{1}{3x}\right)^{15}(23x2​−3x1​)15.

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Q.Find the term independent of $x$ , $x \neq 0$ , in the expansion of $\left(\dfrac{3x^2}{2} - \dfrac{1}{3x}\right)^{15}$ .

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