1 = a1 = a2
an = an -1 + an - 2, n >2
∴ a3 = a2 + a1 = 1 + 1 =2
a4 = a3 + a2 = 2 + 1 = 3
a5 = a4 + a3 = 3 + 2 = 5
a6 = a5 + a4 = 5 + 3 = 8
∴ For n = 1, an + \(\frac{1}{an}=\frac{a2}{a1}=\frac{1}{1}\)
For n = 2 , an + \(\frac{1}{an}=\frac{a3}{a2}=\frac{2}{1}\)
For n = 3 , an + \(\frac{1}{an}=\frac{a4}{a3}=\frac{3}{2}\)
For n = 4 , \(\frac{1}{an}=\frac{a5}{a4}=\frac{5}{3}\)
For n = 5 , an + \(\frac{1}{an}=\frac{a6}{a5}=\frac{8}{5}\).
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