If \(\sum\)\(_{r=1}^n T_r\) = \(\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}\) , then \( \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} \) is equal to :
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
Let $ a_1, a_2, a_3, \ldots $ be in an A.P. such that $$ \sum_{k=1}^{12} 2a_{2k - 1} = \frac{72}{5}, \quad \text{and} \quad \sum_{k=1}^{n} a_k = 0, $$ then $ n $ is:
If $ \theta \in [-2\pi,\ 2\pi] $, then the number of solutions of $$ 2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0 $$ is:
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 $ A = \begin{bmatrix} \alpha & -1 \\6 & \beta \end{bmatrix},\ \alpha > 0 $, such that $ \det(A) = 0 $ and $ \alpha + \beta = 1 $. If $ I $ denotes the $ 2 \times 2 $ identity matrix, then the matrix $ (1 + A)^5 $ is: