Consider the binomial expansion of (p + qx) 9 :
Under what condition the coefficients of x 2 and x 4 are equal ?
Understanding Binomial Expansion and Coefficients
The question asks for the condition under which the coefficients of specific terms in the binomial expansion of \((p + qx)^9\) are equal. To solve this, we need to understand the general term of a binomial expansion.
The binomial theorem states that the expansion of \((a+b)^n\) is given by: \((a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\) where \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\) is the binomial coefficient.
For the given expansion \((p + qx)^9\), we have $a=p$, $b=qx$, and $n=9$. The general term, which is the $(r+1)$-th term, is: \(T_{r+1} = \binom{9}{r} p^{9-r} (qx)^r\) \(T_{r+1} = \binom{9}{r} p^{9-r} q^r x^r\) The coefficient of the term containing \(x^r\) is \(\binom{9}{r} p^{9-r} q^r\).
Finding the Coefficient of \(x^2\)
To find the coefficient of \(x^2\), we need $r=2$. Substituting $r=2$ into the general coefficient formula: \(\text{Coefficient of } x^2 = \binom{9}{2} p^{9-2} q^2 = \binom{9}{2} p^7 q^2\) Now, we calculate the binomial coefficient \(\binom{9}{2}\): \(\binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9!}{2!7!} = \frac{9 \times 8 \times 7!}{2 \times 1 \times 7!} = \frac{9 \times 8}{2} = \frac{72}{2} = 36\) So, the coefficient of \(x^2\) is \(36 p^7 q^2\).
Finding the Coefficient of \(x^4\)
To find the coefficient of \(x^4\), we need $r=4$. Substituting $r=4$ into the general coefficient formula: \(\text{Coefficient of } x^4 = \binom{9}{4} p^{9-4} q^4 = \binom{9}{4} p^5 q^4\) Now, we calculate the binomial coefficient \(\binom{9}{4}\): \(\binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 1 \times 5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126\) So, the coefficient of \(x^4\) is \(126 p^5 q^4\).
Setting the Coefficients Equal
The question states that the coefficients of \(x^2\) and \(x^4\) are equal. Therefore, we set the two coefficients we found equal to each other: \(36 p^7 q^2 = 126 p^5 q^4\)
Solving for the Condition
Now, we need to simplify the equation to find the relationship between $p$ and $q$. Assuming \(p \neq 0\) and \(q \neq 0\) (as they appear in the coefficients), we can divide both sides by common factors.
Divide both sides by \(p^5 q^2\): \(\frac{36 p^7 q^2}{p^5 q^2} = \frac{126 p^5 q^4}{p^5 q^2}\) \(36 p^{7-5} q^{2-2} = 126 p^{5-5} q^{4-2}\) \(36 p^2 q^0 = 126 p^0 q^2\) Since \(q^0 = 1\) and \(p^0 = 1\): \(36 p^2 = 126 q^2\)
Now, we can rearrange this equation to find the ratio of \(p^2\) to \(q^2\). Divide both sides by \(36 q^2\): \(\frac{36 p^2}{36 q^2} = \frac{126 q^2}{36 q^2}\) \(\frac{p^2}{q^2} = \frac{126}{36}\)
Simplify the fraction \(\frac{126}{36}\) by dividing both numerator and denominator by their greatest common divisor, which is 18: \(\frac{126 \div 18}{36 \div 18} = \frac{7}{2}\) So, the condition is: \(\frac{p^2}{q^2} = \frac{7}{2}\) This can be written as a ratio: \(p^2 : q^2 = 7 : 2\)
Checking the Options
Let's compare our derived condition with the given options:
- Option 1: \(p : q = 7 : 2 \implies \frac{p}{q} = \frac{7}{2}\)
- Option 2: \(p^2 : q^2 = 7 : 2 \implies \frac{p^2}{q^2} = \frac{7}{2}\)
- Option 3: \(p : q = 2 : 7 \implies \frac{p}{q} = \frac{2}{7}\)
- Option 4: \(p^2 : q^2 = 2 : 7 \implies \frac{p^2}{q^2} = \frac{2}{7}\)
Our derived condition, \(p^2 : q^2 = 7 : 2\), matches Option 2.
| Term | Value of $r$ | General Term \(T_{r+1}\) | Coefficient |
|---|---|---|---|
| \(x^2\) term | $r=2$ | \(\binom{9}{2} p^{9-2} (qx)^2 = \binom{9}{2} p^7 q^2 x^2\) | \(\binom{9}{2} p^7 q^2 = 36 p^7 q^2\) |
| \(x^4\) term | $r=4$ | \(\binom{9}{4} p^{9-4} (qx)^4 = \binom{9}{4} p^5 q^4 x^4\) | \(\binom{9}{4} p^5 q^4 = 126 p^5 q^4\) |
Revision Table: Binomial Expansion Basics
| Concept | Description | Formula |
|---|---|---|
| Binomial Theorem | Expands a power of a binomial (a+b) into a sum of terms. | \((a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\) |
| General Term | The $(r+1)$-th term in the expansion. | \(T_{r+1} = \binom{n}{r} a^{n-r} b^r\) |
| Binomial Coefficient | The coefficient of each term, represents combinations. | \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\) |
| Coefficient of \(x^k\) in \((a+bx)^n\) | Find $r$ such that \((bx)^r\) gives \(x^k\). This means $r=k$. The coefficient is \(\binom{n}{k} a^{n-k} b^k\). | \(\binom{n}{k} a^{n-k} b^k\) |
Additional Information on Binomial Coefficients
Binomial coefficients, denoted by \(\binom{n}{r}\) or $C(n, r)$, are fundamental in various areas of mathematics, including probability, combinatorics, and algebra.
- They count the number of ways to choose $r$ items from a set of $n$ distinct items without regard to the order of selection.
- They appear as the entries in Pascal's triangle, where each number is the sum of the two numbers directly above it.
- Key properties include \(\binom{n}{r} = \binom{n}{n-r}\), \(\binom{n}{0} = \binom{n}{n} = 1\), and \(\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}\).
Understanding how to calculate and use binomial coefficients is essential for working with binomial expansions and related problems involving combinations.
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