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

Find the term independent of x in the expansion of – 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 $y$ ? …

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To find the term independent of $x$ , we use the general term formula for a binomial expansion, set the exponent of $x$ to zero to find the value of $r$ , and then substitute this $r$ back into the term. The term independent of $x$ is $\boxed{-5{,}674{,}372{,}704}$ .

When we expand an expression like $(a+b)^n$ , the Binomial Theorem gives us a systematic way to find each term. The problem asks for the "term independent of $x$ ," which means we are looking for the term where $x$ does not appear, or more precisely, where the power of $x$ is $0$ (since $x^0 = 1$ ).

The core idea is to first write down the general form of any term in the expansion. This general term will contain $x$ raised to some power involving $r$ (the term index). We then set this power of $x$ to zero and solve for $r$ . Once we have the value of $r$ , we substitute it back into the general term expression to find the specific constant value of that term.

Identify the components of the binomial expression.

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

Comparing this to the standard binomial form $(a+b)^n$ :
- $a = 3x$
- $b = -\dfrac{2}{x^2}$
- $n = 15$
Watch out

Be careful with the sign of $b$ . Here, $b$ is $-\dfrac{2}{x^2}$ , not just $\dfrac{2}{x^2}$ . Forgetting the negative sign is a common mistake.
Write down the general term formula.

The general term, often denoted as $T_{r+1}$ , in the expansion of $(a+b)^n$ is given by:
$T_{r+1} = \binom{n}{r} a^{n-r} b^r$
Substitute the identified components into the general term formula.

Substituting $a=3x$ , $b=-\dfrac{2}{x^2}$ , and $n=15$ :

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

Simplify the general term to isolate the powers of $x$ . We need to separate the constant parts from the $x$ parts.

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

Combine the constant terms and the $x$ terms:

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

Now, combine the powers of $x$ using the rule $x^p \cdot x^q = x^{p+q}$:

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

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

This is the simplified general term. The part $\binom{15}{r} 3^{15-r} (-2)^r$ represents the coefficient, and $x^{15-3r}$ represents the power of $x$.

5. Find the value of $r$ for the term independent of $x$ .

For the term to be independent of $x$ , the exponent of $x$ must be $0$ . …

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Q.Find the term independent of xxx in the expansion of (3x−2x2)15\left(3x - \dfrac{2}{x^2}\right)^{15}(3x−x22​)15.

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

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