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:
\[ nC_2 \cdot (-3)^2 = 10C_2 \cdot 9 = 45 \cdot 9 = 405. \]
\[ \left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \cdots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 \, {}^{14}C_1 \cdots {}^{14}C_{12}} \]
Then \[ 30\alpha = \underline{\hspace{1cm}} \]
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
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
