If the coefficients of $x^7$ in $\left(x^2 + \frac{1}{bx}\right)^{11}$ and $x^{-7}$ in $\left(x - \frac{1}{bx^2}\right)^{11}$, $b \neq 0$, are equal, then the value of b is equal to :
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Remember the general term formula for binomial expansion $(a+b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. Use it to find the power of the variable and solve for 'r' to find the required coefficient. Also, recall the identity $\binom{n}{r} = \binom{n}{n-r}$.
For the first expansion, $\left(x^2 + \frac{1}{bx}\right)^{11}$:
The general term is $T_{r+1} = \binom{11}{r} (x^2)^{11-r} \left(\frac{1}{bx}\right)^r$.
$T_{r+1} = \binom{11}{r} x^{22-2r} \frac{1}{b^r x^r} = \binom{11}{r} \frac{1}{b^r} x^{22-3r}$.
We need the term with $x^7$, so we set the exponent of x to 7:
$22 - 3r = 7 \implies 3r = 15 \implies r = 5$.
The coefficient of $x^7$ is $\binom{11}{5} \frac{1}{b^5}$.
For the second expansion, $\left(x - \frac{1}{bx^2}\right)^{11}$:
The general term is $T_{k+1} = \binom{11}{k} (x)^{11-k} \left(-\frac{1}{bx^2}\right)^k$.
$T_{k+1} = \binom{11}{k} x^{11-k} \frac{(-1)^k}{b^k x^{2k}} = \binom{11}{k} \frac{(-1)^k}{b^k} x^{11-3k}$.
We need the term with $x^{-7}$, so we set the exponent of x to -7:
$11 - 3k = -7 \implies 3k = 18 \implies k = 6$.
The coefficient of $x^{-7}$ is $\binom{11}{6} \frac{(-1)^6}{b^6} = \binom{11}{6} \frac{1}{b^6}$.
We are given that the two coefficients are equal:
$\binom{11}{5} \frac{1}{b^5} = \binom{11}{6} \frac{1}{b^6}$.
Using the property $\binom{n}{r} = \binom{n}{n-r}$, we have $\binom{11}{5} = \binom{11}{11-5} = \binom{11}{6}$.
So, the equation simplifies to:
$\binom{11}{5} \frac{1}{b^5} = \binom{11}{5} \frac{1}{b^6}$.
Since $\binom{11}{5} \neq 0$, we can cancel it from both sides.
$\frac{1}{b^5} = \frac{1}{b^6}$.
$b^6 = b^5$.
$b^6 - b^5 = 0 \implies b^5(b-1) = 0$.
Since we are given $b \neq 0$, the only solution is $b-1=0$, which means $b=1$.
