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

Correct Answer: 3

The general term for the binomial expansion of \( (ax^2 + \frac{70}{27bx})^4 \) is:

\(T_{r+1} = \ ^4C_r (ax^2)^{(4-r)} \times \left( \frac{70}{27bx} \right)^r\)

For the coefficient of \( x^5 \), we need to find \( r \) such that the power of \( x \) equals 5. The power of \( x \) in the general term is:

\(2(4 - r) - r = 5\)

Simplifying:

\(8 - 2r - r = 5 \Rightarrow r = 1\)

Now, substituting \( r = 1 \) into the general term, we get the coefficient of \( x^5 \):

\(\text{Coefficient of } x^5 = \ ^4C_1 a^3 \left( \frac{70}{27b} \right)\)

Next, for the coefficient of \( x^{-5} \), the general term for the binomial expansion of \( (ax + -\frac{1}{bx^2})^7 \) is:

\(t_{r+1} = ^7C_r (ax)^{7-r} \left( -\frac{1}{bx^2} \right)^r\)

For the coefficient of \( x^{-5} \), we solve for \( r \) such that the power of \( x \) equals -5. The power of \( x \) in this term is:

\(7 - r - 2r = -5 \Rightarrow r = 4\)

Now, substituting \( r = 4 \) into the general term, we get the coefficient of \( x^{-5} \):

\(\text{Coefficient of } x^{-5} = \ ^7C_4 a^3 \frac{1}{b^4}\)

We are given that:

\(\ ^7C_4 a^3 \frac{1}{b^4} \Rightarrow 2b = 3\)

Thus, the value of \( 2b \) is 3.