The general term in the binomial expansion of \( (1 + 2x)^n \) is given by: \[ T_k = \binom{n}{k} (2x)^k = \binom{n}{k} 2^k x^k. \] Thus, the coefficient of \( x^k \) is \( \binom{n}{k} 2^k \).
Step 1: Expressing the ratio of the coefficients. Let the three consecutive terms be \( T_r, T_{r+1}, T_{r+2} \), and their coefficients be \( \binom{n}{r} 2^r, \binom{n}{r+1} 2^{r+1}, \binom{n}{r+2} 2^{r+2} \), respectively. The ratio of these coefficients is given by: \[ \frac{\binom{n}{r+1} 2^{r+1}}{\binom{n}{r} 2^r} = \frac{5}{2}, \quad \frac{\binom{n}{r+2} 2^{r+2}}{\binom{n}{r+1} 2^{r+1}} = \frac{8}{5}. \]
Step 2: Simplifying the ratios. Simplifying the first ratio: \[ \frac{\binom{n}{r+1}}{\binom{n}{r}} \cdot 2 = \frac{5}{2} \quad \Rightarrow \quad \frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{5}{4}. \] Using the property of binomial coefficients \( \frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1} \), we get: \[ \frac{n-r}{r+1} = \frac{5}{4} \quad \Rightarrow \quad 4(n-r) = 5(r+1) \quad \Rightarrow \quad 4n - 4r = 5r + 5 \quad \Rightarrow \quad 4n = 9r + 5. \] Simplifying the second ratio: \[ \frac{\binom{n}{r+2}}{\binom{n}{r+1}} \cdot 2 = \frac{8}{5} \quad \Rightarrow \quad \frac{\binom{n}{r+2}}{\binom{n}{r+1}} = \frac{4}{5}. \] Using the property \( \frac{\binom{n}{r+2}}{\binom{n}{r+1}} = \frac{n-r-1}{r+2} \), we get: \[ \frac{n-r-1}{r+2} = \frac{4}{5} \quad \Rightarrow \quad 5(n-r-1) = 4(r+2) \quad \Rightarrow \quad 5n - 5r - 5 = 4r + 8 \quad \Rightarrow \quad 5n = 9r + 13. \]
Step 3: Solving the system of equations. We now have the system of equations: \[ 4n = 9r + 5 \quad \text{and} \quad 5n = 9r + 13. \] Subtract the first equation from the second: \[ 5n - 4n = (9r + 13) - (9r + 5) \quad \Rightarrow \quad n = 8. \]
Step 4: Finding the middle term's coefficient. Substitute \( n = 8 \) into the expression for the coefficient of the middle term: \[ T_{r+1} = \binom{8}{r+1} 2^{r+1}. \] Since \( r = 3 \) (from solving the system of equations), we substitute \( r = 3 \) into the expression for \( T_4 \): \[ T_4 = \binom{8}{4} 2^4 = \binom{8}{4} \cdot 16 = 70 \cdot 16 = 1120. \]
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,