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Step-by-Step Solution
Step 1: Write the General Term of the First Expansion
We want the coefficient of $x^7$ in
$ \left(x^2 + \frac{1}{b x}\right)^{11} $. The general term in the binomial expansion
$ (A + B)^{n} $ is given by
$ {}^nC_r \, A^{n-r} \, B^r $.
Here, let $A = x^2$ and $B = \frac{1}{b x}$. So, the general term (the $(r+1)$-th term) would be:
$$ {}^{11}C_{r} \left(x^2\right)^{11-r} \left(\frac{1}{b x}\right)^r. $$
Step 2: Simplify and Identify the Power of x
The expression simplifies to:
$$ {}^{11}C_r \, x^{2(11-r)} \, \frac{1}{b^r} \, x^{-r}
\;=\; {}^{11}C_r \, \frac{1}{b^r}\, x^{22 - 2r - r}
\;=\; {}^{11}C_r \, \frac{1}{b^r}\, x^{22 - 3r}. $$
We need the exponent of $x$ to be 7. Hence, set
$$ 22 - 3r = 7. $$
Step 3: Solve for r in the First Expansion
Solving the equation:
$$ 22 - 3r = 7 \quad\Longrightarrow\quad 22 - 7 = 3r \quad\Longrightarrow\quad 15 = 3r \quad\Longrightarrow\quad r = 5. $$
Therefore, the term in $x^7$ corresponds to $r = 5$.
Step 4: Extract the Coefficient of x^7
When $r=5$, the coefficient of $x^7$ becomes:
$$ {}^{11}C_5 \,\frac{1}{b^5}. $$
Step 5: Write the General Term of the Second Expansion
We now look at the coefficient of $x^{-7}$ in
$ \left(x - \frac{1}{b x^2}\right)^{11} $. Again, the general term of $(A + B)^{n}$ is
$ {}^nC_r \, A^{n-r} \, B^r $.
Let $A = x$ and $B = -\frac{1}{b x^2}$. Then, the general term is:
$$ {}^{11}C_r \, x^{11-r} \left(-\frac{1}{b x^2}\right)^r. $$
Step 6: Simplify and Identify the Power of x
We simplify:
$$ {}^{11}C_r \, x^{11-r} \, \frac{(-1)^r}{b^r}\, x^{-2r}
= {}^{11}C_r \frac{(-1)^r}{b^r} x^{11 - r - 2r}
= {}^{11}C_r \frac{(-1)^r}{b^r} x^{11 - 3r}. $$
We want the exponent of $x$ to be $-7$. Thus, set
$$ 11 - 3r = -7. $$
Step 7: Solve for r in the Second Expansion
Solving:
$$ 11 - 3r = -7 \quad\Longrightarrow\quad 11 + 7 = 3r \quad\Longrightarrow\quad 18 = 3r \quad\Longrightarrow\quad r = 6. $$
Hence, the term in $x^{-7}$ corresponds to $r = 6$.
Step 8: Extract the Coefficient of x−7
When $r=6$, the coefficient is:
$$ {}^{11}C_6 \frac{(-1)^6}{b^6} = {}^{11}C_6 \frac{1}{b^6}, $$
(noting that $(-1)^6 = 1$).
Step 9: Equate the Two Coefficients
We are given that these two coefficients are equal. Hence:
$$ {}^{11}C_5 \frac{1}{b^5} = {}^{11}C_6 \frac{1}{b^6}. $$
Since $b \neq 0$, we can multiply both sides by $b^6$:
$$ {}^{11}C_5 \, b = {}^{11}C_6. $$
Step 10: Relation Between Binomial Coefficients
Recall that
$$ {}^{n}C_{r} = {}^{n}C_{n-r}. $$
Hence, for $n = 11$,
$$ {}^{11}C_5 = {}^{11}C_{6}. $$
Thus,
$$ {}^{11}C_5 = {}^{11}C_6. $$
This shows that if we substitute back into
$ {}^{11}C_5 \, b = {}^{11}C_6, $
we get
$$ {}^{11}C_5 \, b = {}^{11}C_5 \quad \Longrightarrow\quad b = 1. $$
Conclusion
Therefore, the value of $b$ that satisfies the given condition is
$$ b = 1. $$
