To solve the problem, we need to understand how to compute the sums \( A \) and \( B \). These sums represent the total of all coefficients in the respective expansions. Let's break down the solution step-by-step:
After these steps, it's evident that the relationship between \( A \) and \( B \) is expressed by the formula \( A = B^3 \). Hence, the correct answer is:
\( A = B^3 \)
To find the sums \( A \) and \( B \), we calculate the sum of all coefficients by setting \( x = 1 \) in each expansion.
Step 1. Calculate \( A \)
Substitute \( x = 1 \) in \( (1 - 3x + 10x^2)^n \):\[ A = (1 - 3 \cdot 1 + 10 \cdot 1^2)^n = (1 - 3 + 10)^n = 8^n \]Therefore, \( A = 8^n \).
Step 2. Calculate \( B \)
Substitute \( x = 1 \) in \( (1 + x^2)^n \):\[ B = (1 + 1^2)^n = 2^n \]Thus, \( B = 2^n \).
Step 3. Find the Relationship Between \( A \) and \( B \)
Since \( A = 8^n \) and \( B = 2^n \), we can write:\[ A = (2^n)^3 = B^3 \]Therefore, \( A = B^3 \).
\[ \left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \cdots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 \, {}^{14}C_1 \cdots {}^{14}C_{12}} \]
Then \[ 30\alpha = \underline{\hspace{1cm}} \]
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).

Method used for separation of mixture of products (B and C) obtained in the following reaction is: 