an =\(\frac {n}{n} + 1\)
Substituting n = 1, 2, 3, 4, 5, we obtain
a1 = \(\frac{1}{1} + 1 = \frac{1}{2}\)
a2 = \(\frac{2}{2} + 1 = \frac{2}{3}\)
a3 = \(\frac{3}{3} + 1 = \frac{3}{4}\)
a4 = \(\frac{4}{4} + 1 = \frac{4}{5}\)
a5 = \(\frac{5}{5} + 1 = \frac{5}{6}\)
Therefore, the required terms are \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, and \frac {5}{6}\).
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:
The sum $ 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + ... $ upto $ \infty $ terms, is equal to