Consider the binomial expansion of (p + qx) 9 :
What is the value of q if the coefficients of x 3 and x 6 are equal ?
p
Understanding the Binomial Expansion of (p + qx)9
The problem asks us to find the value of 'q' in the binomial expansion of \((p + qx)^9\), given that the coefficients of the \(x^3\) term and the \(x^6\) term are equal.
Let's start by recalling the binomial theorem, which provides a formula for expanding expressions of the form \((a+b)^n\). The general term (or the \((r+1)\)-th term) in the expansion of \((a+b)^n\) is given by:
\(\text{T}_{r+1} = \binom{n}{r} a^{n-r} b^r\)
In our case, the expression is \((p + qx)^9\). Comparing this with \((a+b)^n\), we have:
- \(a = p\)
- \(b = qx\)
- \(n = 9\)
So, the general term in the expansion of \((p + qx)^9\) is:
\(\text{T}_{r+1} = \binom{9}{r} p^{9-r} (qx)^r\)
We can simplify the term \((qx)^r\) as \(q^r x^r\). Thus, the general term becomes:
\(\text{T}_{r+1} = \binom{9}{r} p^{9-r} q^r x^r\)
The coefficient of the term with \(x^r\) is \(\binom{9}{r} p^{9-r} q^r\).
Finding the Coefficient of x3
To find the term with \(x^3\), we need the exponent of \(x\) in the general term to be 3. From the general term formula \(\binom{9}{r} p^{9-r} q^r x^r\), the exponent of \(x\) is \(r\). So, we set \(r = 3\).
The coefficient of \(x^3\) is the coefficient from the general term when \(r=3\):
\(\text{Coefficient of } x^3 = \binom{9}{3} p^{9-3} q^3\)
\(\text{Coefficient of } x^3 = \binom{9}{3} p^6 q^3\)
Let's calculate the binomial coefficient \(\binom{9}{3}\):
\(\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84\)
So, the coefficient of \(x^3\) is \(84 p^6 q^3\).
Finding the Coefficient of x6
To find the term with \(x^6\), we set \(r = 6\) in the general term formula \(\binom{9}{r} p^{9-r} q^r x^r\).
The coefficient of \(x^6\) is the coefficient from the general term when \(r=6\):
\(\text{Coefficient of } x^6 = \binom{9}{6} p^{9-6} q^6\)
\(\text{Coefficient of } x^6 = \binom{9}{6} p^3 q^6\)
Let's calculate the binomial coefficient \(\binom{9}{6}\):
\(\binom{9}{6} = \frac{9!}{6!(9-6)!} = \frac{9!}{6!3!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84\)
Alternatively, using the property \(\binom{n}{k} = \binom{n}{n-k}\), we have \(\binom{9}{6} = \binom{9}{9-6} = \binom{9}{3} = 84\).
So, the coefficient of \(x^6\) is \(84 p^3 q^6\).
Setting Coefficients Equal and Solving for q
The problem states that the coefficients of \(x^3\) and \(x^6\) are equal. Therefore, we can set the two coefficients we found equal to each other:
\(84 p^6 q^3 = 84 p^3 q^6\)
Assuming \(p \neq 0\) and \(q \neq 0\) (as if \(p=0\) or \(q=0\), the coefficients would likely be zero or the problem trivial), we can divide both sides by common factors. Divide both sides by 84:
\(p^6 q^3 = p^3 q^6\)
Now, divide both sides by \(p^3 q^3\):
\(\frac{p^6 q^3}{p^3 q^3} = \frac{p^3 q^6}{p^3 q^3}\)
Using the exponent rule \(\frac{a^m}{a^n} = a^{m-n}\):
\(p^{6-3} = q^{6-3}\)
\(p^3 = q^3\)
Taking the cube root of both sides gives:
\(\sqrt[3]{p^3} = \sqrt[3]{q^3}\)
\(p = q\)
Thus, the value of \(q\) is equal to \(p\).
Summary of Steps
- Identify the general term formula for the binomial expansion \((p + qx)^9\).
- Use the general term to find the coefficient of \(x^3\) by setting the exponent of \(x\) to 3.
- Use the general term to find the coefficient of \(x^6\) by setting the exponent of \(x\) to 6.
- Set the two coefficients equal to each other as per the problem statement.
- Solve the resulting equation for \(q\) in terms of \(p\).
Conclusion on Finding q
Based on the calculations, for the coefficients of \(x^3\) and \(x^6\) in the binomial expansion of \((p + qx)^9\) to be equal, the value of \(q\) must be equal to \(p\).
| Term | r value | General Term (simplified) | Coefficient |
|---|---|---|---|
| \(x^3\) | 3 | \(\binom{9}{3} p^{6} q^3 x^3\) | \(\binom{9}{3} p^{6} q^3 = 84 p^6 q^3\) |
| \(x^6\) | 6 | \(\binom{9}{6} p^{3} q^6 x^6\) | \(\binom{9}{6} p^{3} q^6 = 84 p^3 q^6\) |
Equating coefficients:
\(84 p^6 q^3 = 84 p^3 q^6\)
\(p^6 q^3 = p^3 q^6\)
\(p^3 = q^3\) (assuming \(p, q \neq 0\))
\(p = q\)
Revision Table: Key Concepts for Binomial Expansion
| Concept | Description | Formula/Example |
|---|---|---|
| Binomial Theorem | Expands powers of a binomial \((a+b)^n\) into a sum of terms. | \((a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\) |
| Binomial Coefficient \(\binom{n}{r}\) | Represents the number of ways to choose \(r\) items from a set of \(n\) items (n choose r). Used as coefficients in the binomial expansion. | \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\) |
| General Term | The \((r+1)\)-th term in the expansion of \((a+b)^n\). | \(\text{T}_{r+1} = \binom{n}{r} a^{n-r} b^r\) |
Additional Information: Properties of Binomial Coefficients
Binomial coefficients \(\binom{n}{r}\) have several useful properties:
- Symmetry: \(\binom{n}{r} = \binom{n}{n-r}\). This property was used when calculating \(\binom{9}{6}\) as \(\binom{9}{3}\).
- Sum of coefficients: The sum of the binomial coefficients in the expansion of \((a+b)^n\) is \(2^n\). This is found by setting \(a=1\) and \(b=1\) in the binomial theorem.
- Pascal's Identity: \(\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}\). This identity is the basis for constructing Pascal's triangle.
Understanding these properties can simplify calculations and help solve problems related to binomial expansions.
Important Questions from Binomial Expansion
If $x = \frac{1}{4}$, then the greatest term in the expansion of $(2 + 3x)^{15}$ will be
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