If the coefficient of `x^(8)` in `(ax^(2) + (1)/( bx) )^(13)` is equal to the coefficient of `x^(-8)` in `(ax- (1)/( bx^(2) ) )^(13)`, then a and b will satisfy the relation

Options

A. `ab+1=0`
B. `ab=1`
C. `a=1-b`
D. `a+b=-1`

Answer

To solve the problem, we need to find the coefficients of \(x^8\) and \(x^{-8}\) in the given expressions and set them equal to each other.

### Step 1: Find the coefficient of \(x^8\) in \((ax^2 + \frac{1}{bx})^{13}\)

The general term (T) in the expansion of \((a x^2 + \frac{1}{b x})^{13}\) is given by:

\[ T_{r+1} = \binom{13}{r} (a x^2)^{13-r} \left(\frac{1}{b x}\right)^r \]

This simplifies to:

\[ T_{r+1} = \binom{13}{r} a^{13-r} \frac{1}{b^r} x^{2(13-r) - r} \]

Now, we need the exponent of \(x\) to be 8:

\[ 2(13 - r) - r = 8 \]

This simplifies to:

\[ 26 - 2r - r = 8 \implies 26 - 3r = 8 \implies 3r = 18 \implies r = 6 \]

Substituting \(r = 6\) into the general term gives us:

\[ T_{7} = \binom{13}{6} a^{7} \frac{1}{b^6} x^{8} \]

Thus, the coefficient of \(x^8\) is:

\[ \text{Coefficient of } x^8 = \binom{13}{6} a^{7} \frac{1}{b^6} \]

### Step 2: Find the coefficient of \(x^{-8}\) in \((ax - \frac{1}{bx^2})^{13}\)

The general term in the expansion of \((ax - \frac{1}{bx^2})^{13}\) is:

\[ T_{r+1} = \binom{13}{r} (ax)^{13-r} \left(-\frac{1}{b x^2}\right)^r \]

This simplifies to:

\[ T_{r+1} = \binom{13}{r} a^{13-r} (-1)^r \frac{1}{b^r} x^{13 - r - 2r} = \binom{13}{r} a^{13-r} (-1)^r \frac{1}{b^r} x^{13 - 3r} \]

We need the exponent of \(x\) to be \(-8\):

\[ 13 - 3r = -8 \]

This simplifies to:

\[ 13 + 8 = 3r \implies 21 = 3r \implies r = 7 \]

Substituting \(r = 7\) into the general term gives us:

\[ T_{8} = \binom{13}{7} a^{6} (-1)^7 \frac{1}{b^7} x^{-8} \]

Thus, the coefficient of \(x^{-8}\) is:

\[ \text{Coefficient of } x^{-8} = -\binom{13}{7} a^{6} \frac{1}{b^7} \]

### Step 3: Set the coefficients equal to each other

Now we set the coefficients equal to each other:

\[ \binom{13}{6} a^{7} \frac{1}{b^6} = -\binom{13}{7} a^{6} \frac{1}{b^7} \]

### Step 4: Simplify the equation

Cancelling out common terms:

\[ \binom{13}{6} a^{7} b = -\binom{13}{7} a^{6} \]

Dividing both sides by \(a^{6}\) (assuming \(a \neq 0\)):

\[ \binom{13}{6} a b = -\binom{13}{7} \]

### Step 5: Use the identity of binomial coefficients

Using the identity \(\binom{n}{r} = \frac{n-r}{r+1} \binom{n}{r-1}\):

\[ \binom{13}{7} = \binom{13}{6} \frac{6}{7} \]

Substituting this into our equation gives:

\[ \binom{13}{6} a b = -\binom{13}{6} \frac{6}{7} \]

Cancelling \(\binom{13}{6}\) (assuming it is non-zero):

\[ a b = -\frac{6}{7} \]

### Conclusion

Thus, the relation that \(a\) and \(b\) satisfy is:

\[ ab + \frac{6}{7} = 0 \quad \text{or} \quad ab = -\frac{6}{7} \]