Understanding Binomial Expansion and Term Calculation

The problem asks us to find the ratio \(\frac{x}{y}\) given information about specific terms in the binomial expansion of \((x - y)^n\). The expansion is done in descending powers of x, and we are told that the sum of the fifth and sixth terms is zero.

Let's first recall the general term in the binomial expansion of \((a+b)^n\). The \((r+1)^{\text{th}}\) term is given by the formula:

\(T_{r+1} = \binom{n}{r} a^{n-r} b^r\)

In our case, the expression is \((x - y)^n\), which can be written as \((x + (-y))^n\). Here, \(a = x\) and \(b = -y\). The general term \(T_{r+1}\) in the expansion of \((x - y)^n\) is therefore:

\(T_{r+1} = \binom{n}{r} x^{n-r} (-y)^r = \binom{n}{r} x^{n-r} (-1)^r y^r\)

Calculating the Fifth and Sixth Terms

We need to find the fifth term (\(T_5\)) and the sixth term (\(T_6\)).

• For the fifth term, we set \(r+1 = 5\), which means \(r = 4\).

The fifth term is:

\(T_5 = T_{4+1} = \binom{n}{4} x^{n-4} (-y)^4 = \binom{n}{4} x^{n-4} y^4\)

(Since \((-y)^4 = (-1)^4 y^4 = 1 \cdot y^4 = y^4\))

• For the sixth term, we set \(r+1 = 6\), which means \(r = 5\).

The sixth term is:

\(T_6 = T_{5+1} = \binom{n}{5} x^{n-5} (-y)^5 = \binom{n}{5} x^{n-5} (-1)^5 y^5 = -\binom{n}{5} x^{n-5} y^5\)

(Since \((-y)^5 = (-1)^5 y^5 = -1 \cdot y^5 = -y^5\))

Using the Condition: Sum of Terms is Zero

We are given that the sum of the fifth term and the sixth term is zero:

\(T_5 + T_6 = 0\)

Substitute the expressions we found for \(T_5\) and \(T_6\):

\(\binom{n}{4} x^{n-4} y^4 + \left(-\binom{n}{5} x^{n-5} y^5\right) = 0\)

\(\binom{n}{4} x^{n-4} y^4 = \binom{n}{5} x^{n-5} y^5\)

Solving for the Ratio \(\frac{x}{y}\)

Now, we need to rearrange the equation to isolate \(\frac{x}{y}\). We can divide both sides by \(\binom{n}{5} x^{n-5} y^4\) (assuming \(x, y \ne 0\), which is implied if terms exist and sum to zero and \(n \ge 5\)):

\(\frac{\binom{n}{4}}{\binom{n}{5}} = \frac{x^{n-5} y^5}{x^{n-4} y^4}\)

Let's simplify the combination terms using the formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) and \(\binom{n}{k} / \binom{n}{k-1} = \frac{n-k+1}{k}\). In our case, \(\binom{n}{5} / \binom{n}{4} = \frac{n-5+1}{5} = \frac{n-4}{5}\). So, \(\frac{\binom{n}{4}}{\binom{n}{5}} = \frac{5}{n-4}\).

Alternatively, using factorials:

\(\frac{\binom{n}{4}}{\binom{n}{5}} = \frac{\frac{n!}{4!(n-4)!}}{\frac{n!}{5!(n-5)!}} = \frac{n!}{4!(n-4)!} \times \frac{5!(n-5)!}{n!}\)

Cancel out \(n!\):

\(= \frac{5!(n-5)!}{4!(n-4)!}\)

Expand factorials: \(5! = 5 \times 4!\) and \((n-4)! = (n-4) \times (n-5)!\)

\(= \frac{5 \times 4! \times (n-5)!}{4! \times (n-4) \times (n-5)!}\)

Cancel out \(4!\) and \((n-5)!\):

\(= \frac{5}{n-4}\)

Now, let's simplify the terms involving x and y:

\(\frac{x^{n-5} y^5}{x^{n-4} y^4} = x^{n-5 - (n-4)} y^{5-4} = x^{n-5 - n + 4} y^1 = x^{-1} y = \frac{y}{x}\)

So, the equation becomes:

\(\frac{5}{n-4} = \frac{y}{x}\)

We need to find \(\frac{x}{y}\), so we take the reciprocal of both sides:

\(\frac{x}{y} = \frac{n-4}{5}\)

Comparing with Options

Let's compare our result \(\frac{x}{y} = \frac{n-4}{5}\) with the given options:

1. \(\frac{{{\rm{n}} - 5}}{6}\)
2. \(\frac{{{\rm{n}} - 4}}{5}\)
3. \(\frac{5}{{{\rm{n}} - 4}}\)
4. \(\frac{6}{{{\rm{n}} - 5}}\)

Our derived ratio matches option 2.

Revision Table: Binomial Expansion Concepts

Additional Information: Solving Binomial Problems

When tackling problems involving specific terms in a binomial expansion like \((x-y)^n\), it's crucial to correctly identify the components 'a' and 'b' in the general \((a+b)^n\) formula. Remember that the sign is part of the 'b' term. In \((x-y)^n\), 'a' is \(x\) and 'b' is \(-y\).

The power of \(-y\) in the \((r+1)^{\text{th}}\) term is \(r\). This means the term will have a sign determined by \((-1)^r\). If \(r\) is even, the term is positive; if \(r\) is odd, the term is negative.

Problems where the sum of terms is zero often lead to an equation where the terms are equal in magnitude but opposite in sign. This allows you to set up an equation involving the binomial coefficients and the variables (like x and y in this case) and solve for the required ratio.

Understanding how to manipulate factorial expressions and binomial coefficients is key to simplifying such equations efficiently. The property \(\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}\) is particularly useful for ratios of consecutive terms.