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 \[ 30\alpha = \underline{\hspace{1cm}} \]
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]