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:
\[ 2b = 3 \]
The value of \( 2b \) is \( \boxed{3} \).
\[ \left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \cdots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 \, {}^{14}C_1 \cdots {}^{14}C_{12}} \]
Then \[ 30\alpha = \underline{\hspace{1cm}} \]
Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
