Understanding Binomial Expansion Middle Term

The question asks to find the middle term of the binomial expansion of the expression \(\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^8\).

The general form of a binomial expansion is \((a+b)^n\). In this case, we have:

• \(a = \dfrac{x}{y}\)
• \(b = \dfrac{y}{x}\)
• \(n = 8\)

When the power \(n\) of a binomial expansion is even, there is a single middle term. The position of the middle term is given by the formula \(\left(\dfrac{n}{2} + 1\right)^{th}\) term.

For the given expression, \(n=8\). Plugging this into the formula:

Middle term position = \(\left(\dfrac{8}{2} + 1\right)^{th} = (4 + 1)^{th} = 5^{th}\) term.

Calculating the Middle Term

The general formula for the \((k+1)^{th}\) term in the binomial expansion of \((a+b)^n\) is:

$$T_{k+1} = \binom{n}{k} a^{n-k} b^k$$

To find the \(5^{th}\) term, we need \(k+1 = 5\), which means \(k=4\).

Now, substitute the values of \(n\), \(k\), \(a\), and \(b\) into the general term formula:

• \(n = 8\)
• \(k = 4\)
• \(a = \dfrac{x}{y}\)
• \(b = \dfrac{y}{x}\)

$$T_{4+1} = T_5 = \binom{8}{4} \left(\dfrac{x}{y}\right)^{8-4} \left(\dfrac{y}{x}\right)^4$$

Simplify the expression:

$$T_5 = \binom{8}{4} \left(\dfrac{x}{y}\right)^{4} \left(\dfrac{y}{x}\right)^4$$

$$T_5 = \binom{8}{4} \left(\dfrac{x^4}{y^4}\right) \left(\dfrac{y^4}{x^4}\right)$$

Notice that \(\dfrac{x^4}{y^4} \times \dfrac{y^4}{x^4} = 1\).

$$T_5 = \binom{8}{4} \times 1$$

$$T_5 = \binom{8}{4}$$

Identifying the Correct Option

The calculated middle term is \(\binom{8}{4}\).

Comparing this with the given options:

• Option 1: \({8_{{C_4}}}\) which represents \(\binom{8}{4}\)
• Option 2: \({8_{{C_5}}}\) which represents \(\binom{8}{5}\)
• Option 3: \({8_{{C_6}}}\) which represents \(\binom{8}{6}\)
• Option 4: \({8_{{C_7}}}\) which represents \(\binom{8}{7}\)
• Option 5: None of the above

The middle term matches Option 1.