Let the sum of the coefficients of the first three terms in the expansion of $\left(x-\frac{3}{x^2}\right)^n, x \neq 0 n \in N$, be $376$. Then the coefficient of $x^4$ is ______
Step 1: Expand the series
The general term in the expansion of the series is given by:
\[ nC_r \cdot x^{n-r} \cdot (-3)^{2r} = nC_r \cdot (-3)^r \cdot x^{n - 3r}. \]
Step 2: Sum of coefficients of the first three terms
The first three terms correspond to \(r = 0, 1, 2\). These terms are:
\[ T_0 = nC_0 \cdot x^n, \quad T_1 = nC_1 \cdot (-3) \cdot x^{n-3}, \quad T_2 = nC_2 \cdot 9 \cdot x^{n-6}. \]
The sum of the coefficients is:
\[ nC_0 - nC_1 \cdot 3 + nC_2 \cdot 9 = 376. \]
Step 3: Solve for \(n\)
To solve for \(n\), simplify the equation:
\[ 1 - 3n + \frac{n(n - 1)}{2} \cdot 9 = 376. \] Simplifying further: \[ 9n^2 - 27n - 752 = 0. \]
Now, solve the quadratic equation:
\[ n = 10. \]
Step 4: Coefficient of \(x^4\)
To find the coefficient of \(x^4\), we set \(n - 3r = 4\). This gives:
\[ r = \frac{n - 4}{3} = \frac{10 - 4}{3} = 2. \]
The coefficient is:
\[ nC_2 \cdot (-3)^2 = 10C_2 \cdot 9 = 45 \cdot 9 = 405. \]
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,
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
