Show that middle term in the expansion of (x 1x)^2n is 1.3.5....( 2n 1) \ ( 2)^nn! . [6]

Solution

We are to show that the middle term in the expansion of ${\left(x-\frac{1}{x}\right)}^{2n}$ is given by:

$$\frac{1\cdot 3\cdot 5\cdots (2n-1)\cdot (-2{)}^n}{n!}$$

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Step 1: General Expansion Using Binomial Theorem

The binomial expansion of ${\left(x-\frac{1}{x}\right)}^{2n}$ is given by:

$${\left(x-\frac{1}{x}\right)}^{2n}=\sum _{k=0}^{2n}\left(\genfrac{}{}{0ex}{}{2n}{k}\right)x^{2n-k}{\left(-\frac{1}{x}\right)}^k$$

Simplifying the exponents:

$$=\sum _{k=0}^{2n}\left(\genfrac{}{}{0ex}{}{2n}{k}\right)(-1{)}^k{x}^{2n-2k}$$

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Step 2: Identifying the Middle Term

Since the exponent is $2n$ (an even number), the expansion has $2n+1$ terms. The middle term corresponds to the term where $k=n$.

Thus, the middle term is:

$$T_{n+1}=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)(-1{)}^n{x}^{2n-2n}=(\genfrac{}{}{0ex}{}{2n}{n})(-1{)}^n{x}^0=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)(-1{)}^n$$

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Step 3: Simplifying the Binomial Coefficient

We need to simplify $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$:

$$\left(\genfrac{}{}{0ex}{}{2n}{n}\right)=\frac{(2n)!}{n!\cdot n!}$$

Now, express $(2n)!$ in terms of double factorials:

$$(2n)!=1\cdot 2\cdot 3\cdot 4\cdots (2n-1)\cdot (2n)$$

Grouping the terms into odd and even factors:

$$(2n)!=[1\cdot 3\cdot 5\cdots (2n-1)]\cdot [2\cdot 4\cdot 6\cdots (2n)]$$

The even terms can be written as:

$$2\cdot 4\cdot 6\cdots (2n)=2^n(1\cdot 2\cdot 3\cdots n)=2^n\cdot n!$$

Thus,

$$(2n)!=[1\cdot 3\cdot 5\cdots (2n-1)]\cdot 2^n\cdot n!$$

Substituting back into the binomial coefficient:

$$\left(\genfrac{}{}{0ex}{}{2n}{n}\right)=\frac{[1\cdot 3\cdot 5\cdots (2n-1)]\cdot 2^n\cdot n!}{n!\cdot n!}=\frac{1\cdot 3\cdot 5\cdots (2n-1)\cdot 2^n}{n!}$$

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Step 4: Substituting Back into the Middle Term

Recall that the middle term is:

$$T_{n+1}=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)(-1{)}^n=\frac{1\cdot 3\cdot 5\cdots (2n-1)\cdot 2^n}{n!}\cdot (-1{)}^n$$

Simplifying the powers of $-1$ and $2$:

$$2^n\cdot (-1{)}^n=(-2{)}^n$$

Thus, the middle term becomes:

$$T_{n+1}=\frac{1\cdot 3\cdot 5\cdots (2n-1)\cdot (-2{)}^n}{n!}$$

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Conclusion

We have shown that the middle term in the expansion of ${\left(x-\frac{1}{x}\right)}^{2n}$ is indeed:

$$\boxed{\frac{1\cdot 3\cdot 5\cdots (2n-1)\cdot (-2{)}^n}{n!}}$$