a1 = -1, an = \(\frac{an-1}{n}\), n ≥ 2
⇒ a2 = \(\frac{a1}{2}=\frac{-1}{2}\)
a3 = \(\frac{a2}{3}=\frac{-1}{6}\)
a4 = \(\frac{a3}{4}=\frac{-1}{24}\)
a5 = \(\frac{a4}{4}=\frac{-1}{120}\)
Hence, the first five terms of the sequence are -1 , \(\frac{-1}{2},\frac{-1}{6},\frac{-1}{24},and\frac{-1}{120}\).
The corresponding series is (-1) + \((\frac{-1}{2})+(\frac{-1}{6})+(\frac{-1}{24})+(\frac{-1}{120})\) + ....
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