Substituting n = 1, 2, 3, 4, 5, we obtain
a1 = 1.12+\(\frac{5}{4}=\frac{6}{4}=\frac{3}{2}\)
a2 = 2.22+\(\frac{5}{4}=\frac{2.9}{4}=\frac{9}{2}\)
a3 = 3.32+\(\frac{5}{4}=\frac{3.14}{4}=\frac{21}{2}\)
a4 = 4.42+\(\frac{5}{4}\) = 21
a5 = 5.52+\(\frac{5}{4}=\frac{5.30}{4}=\frac{75}{2}\)
Therefore, the required terms are \(\frac{3}{2},\frac{9}{2},\frac{21}{2},21,and\frac{75}{2}.\)
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:
The sum $ 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + ... $ upto $ \infty $ terms, is equal to