Show that the middle term in the expansion of $\left(x - \dfrac{1}{x}\right)^{2n}$ is $\dfrac{1 \times 3 \times 5 \times \dots (2n - 1)}{n!} \times (-2)^n$.

> [!TLDR] > The expansion of $\left(x - \dfrac{1}{x}\right)^{2n}$ has $2n+1$ terms, making the $(n+1)$-th term the unique middle term. By applying the binomial theorem's general term formula and simplifying the binomial coefficient $\binom{2n}{n}$, we show the middle term is $\boxed{\dfrac{1 \times 3 \times 5 \times \dots (2n - 1)}{n!} \times (-2)^n}$. The problem asks us to find the middle term in the binomial expansion of $\left(x - \dfrac{1}{x}\right)^{2n}$. To do this, we first need to understand how to identify the middle term and then use the general term formula from the Binomial Theorem. The Binomial Theorem states that for any positive integer $N$, the expansion of $(a+b)^N$ is given by: $$(a+b)^N = \binom{N}{0}a^N b^0 + \binom{N}{1}a^{N-1}b^1 + \dots + \binom{N}{r}a^{N-r}b^r + \dots + \binom{N}{N}a^0 b^N$$ This expansion has $N+1$ terms. > [!FORMULA] > The general term, or $(r+1)$-th term, in the expansion of $(a+b)^N$ is given by: > $T_{r+1} = \binom{N}{r} a^{N-r} b^r$ The number of terms in an expansion $(a+b)^N$ is $N+1$. * If $N$ is even, then $N+1$ is odd. In this case, there is exactly one middle term. Its position is $\left(\frac{N}{2} + 1\right)$-th. * If $N$ is odd, then $N+1$ is even. In this case, there are two middle terms. Their positions are $\left(\frac{N+1}{2}\right)$-th and $\left(\frac{N+1}{2} + 1\right)$-th. In our problem, the power is $2n$. Since $2n$ is always an even number (for any integer $n \ge 1$), the number of terms in the expansion will be $2n+1$, which is an odd number. Therefore, there will be only one middle term. Let's proceed step-by-step to find this middle term. 1. **Determine the position of the middle term.** The power of the binomial is $N = 2n$. The total number of terms in the expansion is $N+1 = 2n+1$. Since the number of terms is odd, there is a single middle term. Its position is given by $\frac{(N+1)+1}{2}$. Substituting $N=2n$: Position of middle term $= \frac{(2n+1)+1}{2} = \fr

Show that the middle term in the expansion of (x - – Class 11 Mathematics

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\left(x - \dfrac{1}{x}\right)^{2n}

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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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The problem requires finding the middle term in the binomial expansion of $\left(x - \dfrac{1}{x}\right)^{2n}$ .

The key idea is using the Binomial Theorem Expansion and the general term formula.

The expansion of $(a+b)^N$ has $N+1$ terms. For $\left(x - \dfrac{1}{x}\right)^{2n}$ , $N=2n$ , so there are $2n+1$ terms. Since $2n+1$ is an odd number, there is a single middle term, which is the $\left(\frac{2n+1+1}{2}\right)$ -th term, i.e., the $(n+1)$ -th term.
The general term in the expansion of $(a+b)^N$ is $T_{r+1} = \binom{N}{r} a^{N-r} b^r$ . Here, $N=2n$ , $a=x$ , $b=-\frac{1}{x}$ , and for the $(n+1)$ -th term, $r=n$ .

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

T_{n+1} = \binom{2n}{n} x^n \frac{(-1)^n}{x^n}

T_{n+1} = \binom{2n}{n} (-1)^n

Now, we simplify the binomial coefficient $\binom{2n}{n}$ :

\binom{2n}{n} = \frac{(2n)!}{n! n!}

We can write $(2n)! = (2n)(2n-1)(2n-2)\dots(4)(3)(2)(1)$.
Separating the even and odd factors:

(2n)! = [(2n)(2n-2)\dots(4)(2)] \times [(2n-1)(2n-3)\dots(3)(1)]

The product of even factors is $2^n (n(n-1)\dots(2)(1)) = 2^n n!$.
So, $(2n)! = 2^n n! \times [1 \times 3 \times 5 \times \dots (2n-1)]$.
Substituting this back into the binomial coefficient: …

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The expansion of $\left(x - \dfrac{1}{x}\right)^{2n}$ has $2n+1$ terms, making the $(n+1)$ -th term the unique middle term. By applying the binomial theorem's general term formula and simplifying the binomial coefficient $\binom{2n}{n}$ , we show the middle term is $\boxed{\dfrac{1 \times 3 \times 5 \times \dots (2n - 1)}{n!} \times (-2)^n}$ .

The problem asks us to find the middle term in the binomial expansion of $\left(x - \dfrac{1}{x}\right)^{2n}$ . To do this, we first need to understand how to identify the middle term and then use the general term formula from the Binomial Theorem.

The Binomial Theorem states that for any positive integer $N$ , the expansion of $(a+b)^N$ is given by:

(a+b)^N = \binom{N}{0}a^N b^0 + \binom{N}{1}a^{N-1}b^1 + \dots + \binom{N}{r}a^{N-r}b^r + \dots + \binom{N}{N}a^0 b^N

This expansion has $N+1$ terms.

(r+1)

The number of terms in an expansion $(a+b)^N$ is $N+1$ .

If $N$ is even, then $N+1$ is odd. In this case, there is exactly one middle term. Its position is $\left(\frac{N}{2} + 1\right)$ -th.
If $N$ is odd, then $N+1$ is even. In this case, there are two middle terms. Their positions are $\left(\frac{N+1}{2}\right)$ -th and $\left(\frac{N+1}{2} + 1\right)$ -th.

In our problem, the power is $2n$ . Since $2n$ is always an even number (for any integer $n \ge 1$ ), the number of terms in the expansion will be $2n+1$ , which is an odd number. Therefore, there will be only one middle term.

Let's proceed step-by-step to find this middle term.

Determine the position of the middle term.

The power of the binomial is $N = 2n$ .

The total number of terms in the expansion is $N+1 = 2n+1$ .

Since the number of terms is odd, there is a single middle term. Its position is given by $\frac{(N+1)+1}{2}$ .

Substituting $N=2n$ :

Position of middle term $= \frac{(2n+1)+1}{2} = \frac{2n+2}{2} = n+1$ .

So, the middle term is the $(n+1)$ -th term, which we denote as $T_{n+1}$ .
Identify the components for the general term formula.

For the expansion $\left(x - \dfrac{1}{x}\right)^{2n}$ :
- $N = 2n$
- $a = x$
- $b = -\dfrac{1}{x}$
- Since we are looking for the $(n+1)$ -th term, we set $r+1 = n+1$ , which means $r=n$ .
Apply the general term formula.

Substitute these values into the formula $T_{r+1} = \binom{N}{r} a^{N-r} b^r$ :

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

Simplify the expression.

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

T_{n+1} = \binom{2n}{n} x^n \frac{(-1)^n}{x^n}

The $x^n$ terms cancel out:

T_{n+1} = \binom{2n}{n} (-1)^n

Expand the binomial coefficient $\binom{2n}{n}$ . The binomial coefficient $\binom{2n}{n}$ is defined as $\frac{(2n)!}{n!n!}$ . We need to manipulate $(2n)!$ to match the desired form.

(2n)! = (2n)(2n-1)(2n-2)\dots(4)(3)(2)(1)

We can separate the even and odd factors:

(2n)! = [(2n)(2n-2)\dots(4)(2)] \times [(2n-1)(2n-3)\dots(3)(1)]

The product of even factors can be written as:

(2n)(2n-2)\dots(4)(2) = 2^n (n)(n-1)\dots(2)(1) = 2^n n!

So, we have: …

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Q.Show that the middle term in the expansion of $\left(x - \dfrac{1}{x}\right)^{2n}$ is $\dfrac{1 \times 3 \times 5 \times \dots (2n - 1)}{n!} \times (-2)^n$ .

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