\(\frac{^nC_4 (2^\frac{1}{4})^{n-4}(3^{\frac{-1}{4}})^4}{^nC_4 (2^\frac{-1}{4})^{n-4}(3^{\frac{1}{4}})^4}\) = \(\sqrt 6\)
\((\frac{2^\frac{1}{4}}{3^\frac{-1}{4}})^{(n-8)}\) = \(\sqrt 6\)
\((6)^\frac{n-8}{4}\) = \(\sqrt 6\)
\(n-8=2\)
\(n=10\)
\(T_3 = {^{10}}C_2(2^{\frac{1}{4}})^8 (3^\frac{-1}{4})^2\)
\(= {^{10}}C_2\times(\sqrt2)^4\times\frac{1}{\sqrt3} = 60\sqrt3\)
So, the correct answer is (B): \(60\sqrt 3\)
If
$ 2^m 3^n 5^k, \text{ where } m, n, k \in \mathbb{N}, \text{ then } m + n + k \text{ is equal to:} $
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 _______
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
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