a1 = 3, an = 3an-1 +2 for all n>1
⇒ a2 = 3a1 + 2 = 3(3) + 2 = 11
a3 = 3a2 +2 = 3(11) + 2 =35
a4 = 3a3+2 = 3(35) + 2 = 107
a5 = 3a4+2 = 3(107) + 2 = 323
Hence, the first five terms of the sequence are 3, 11, 35, 107, and 323.
The corresponding series is 3 + 11 + 35 + 107 + 323 + ….
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