an = n(n+2)
Substituting n = 1, 2, 3, 4, and 5, we obtain
a1= 1(1+2)=3
a2= 2(2+2)=8
a3= 3(3+2)=15
a4= 4(4+2)=24
a5= 5(5+2)=35
Therefore, the required terms are 3, 8, 15, 24, and 35.
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