\(\frac{^nC_4 (2^\frac{1}{4})^{n-4}(3^{\frac{-1}{4}})^4}{^nC_4 (2^\frac{-1}{4})^{n-4}(3^{\frac{1}{4}})^4}\) = \(\sqrt 6\)
\((\frac{2^\frac{1}{4}}{3^\frac{-1}{4}})^{(n-8)}\) = \(\sqrt 6\)
\((6)^\frac{n-8}{4}\) = \(\sqrt 6\)
\(n-8=2\)
\(n=10\)
\(T_3 = {^{10}}C_2(2^{\frac{1}{4}})^8 (3^\frac{-1}{4})^2\)
\(= {^{10}}C_2\times(\sqrt2)^4\times\frac{1}{\sqrt3} = 60\sqrt3\)
So, the correct answer is (B): \(60\sqrt 3\)
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is:
Let $ (1 + x + x^2)^{10} = a_0 + a_1 x + a_2 x^2 + ... + a_{20} x^{20} $. If $ (a_1 + a_3 + a_5 + ... + a_{19}) - 11a_2 = 121k $, then k is equal to _______
For \( \alpha, \beta, \gamma \in \mathbb{R} \), if \[ \lim_{x \to 0} \frac{x^2 \sin(\alpha x) + (\gamma - 1)e^{x^2}}{\sin(2x - \beta x)} = 3, \] then \( \beta + \gamma - \alpha \) is equal to:

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
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
