The sum $ 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + ... $ upto $ \infty $ terms, is equal to
If $ \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + ... \infty = \frac{\pi^4}{90}, $ $ \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + ... \infty = \alpha, $ $ \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + ... \infty = \beta, $ then $ \frac{\alpha}{\beta} $ is equal to:
Let \( a_1, a_2, a_3, \ldots \) be in an A.P. such that \[ \sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1, \quad a_1 \neq 0. \] If \[ \sum_{k=1}^{n} a_k = 0, \] then \( n \) is: