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

Answer 1

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 \( \boxed{3} \).