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: Dec 16, 2024
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Solution and Explanation
\(T_{r+1}=\ ^4C_r(ax^2)^{(4-r)}\times(\frac{70}{27bx})^r\)
For coefficient of \(x^5\),\(8-2r-r=5\Rightarrow r=1\)
\(\Rightarrow\) Coefficient of \(x^5 = \ ^4C_1a^3(\frac{70}{27b})\)
\(t_{r+1}=^7C_r(ax)^{7-r}(-\frac{1}{bx^2})^r\)
for the coefficient of x-5, \(7-r-2r=-5\Rightarrow r=4\)
Coefficient of x-5 = \(^7C_4a^3\frac{1}{b^4}\Rightarrow 2b=3\)
The value of 2b is 3.
Therefore, the correct answer is 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.