Substituting n = 1, 2, 3, 4, 5, we obtain
a1 = (-1)1-1 51+1 = 52 = 25
a2 = (-1)2-1 52+1 = -53 = -125
a3 = (-1)3-1 53+1 = -54 = 625
a4 = (-1)4-1 54+1 = -55 = -3125
a5 = (-1)5-1 55+1 = 56 = 15625
Therefore, the required terms are 25, –125, 625, –3125, and 15625.
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