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, \( A = B^3 \).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,