Question:
Let a and b be two nonzero real numbers. If the coefficient of x5 in the expansion of (\(ax^2+(\frac{70}{27bx})^4\) is equal to the coefficient of x-5 in the expansion of \((ax-\frac{1}{bx^2})^7\). then the value of 2b is
Let a and b be two nonzero real numbers. If the coefficient of x5 in the expansion of (\(ax^2+(\frac{70}{27bx})^4\) is equal to the coefficient of x-5 in the expansion of \((ax-\frac{1}{bx^2})^7\). then the value of 2b is
Updated On: May 8, 2025
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Solution and Explanation
We are given the expression:
\[ T_{r+1} = \ ^4C_r (ax^2)^{(4-r)} \times \left(\frac{70}{27bx}\right)^r \]
For the coefficient of \( x^5 \), we solve for \( r \):
\[ 8 - 2r - r = 5 \quad \Rightarrow \quad r = 1 \]
Thus, the coefficient of \( x^5 \) is:
\[ \ ^4C_1 a^3 \left( \frac{70}{27b} \right) \]
Next, for the expression:
\[ t_{r+1} = \ ^7C_r (ax)^{7-r} \left( -\frac{1}{bx^2} \right)^r \]
For the coefficient of \( x^{-5} \), we solve for \( r \):
\[ 7 - r - 2r = -5 \quad \Rightarrow \quad r = 4 \]
The coefficient of \( x^{-5} \) is:
\[ \ ^7C_4 a^3 \frac{1}{b^4} \]
Thus, we have the equation:
\[ 2b = 3 \]
Final Answer:
The value of \( 2b \) is \( \boxed{3} \).
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Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
- The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
- There are (n+1) terms in the expansion of (x+y)n.
- The first and the last terms are xn and yn respectively.
- From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.