a1 = a2 = 2, an = an-1 -1, n > 2
⇒ a3 = a2 - 1 = 2 - 1 = 1
a4 = a3 - 1 = 1-1 = 0
a5 = a4 - 1 = 0 - 1 = -1 Hence, the first five terms of the sequence are 2, 2, 1, 0, and –1.
The corresponding series is 2 + 2 + 1 + 0 + (–1) + …
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