Step 1: Determine the number of unpaired electrons for each complex using Crystal Field Theory.
\( [Co(NH_3)_6]^{3+} \) (\( d^6 \), strong field): 0 unpaired electrons (diamagnetic)
\( [Co(C_2O_4)_3]^{3-} \) (\( d^6 \), weak field): 4 unpaired electrons (paramagnetic)
\( [MnCl_6]^{3-} \) (\( d^4 \), weak field): 4 unpaired electrons (paramagnetic)
\( [Mn(CN)_6]^{3-} \) (\( d^4 \), strong field): 2 unpaired electrons (paramagnetic)
\( [CoF_6]^{3-} \) (\( d^6 \), weak field): 4 unpaired electrons (paramagnetic)
\( [Fe(CN)_6]^{3-} \) (\( d^5 \), strong field): 1 unpaired electron (paramagnetic)
\( [FeF_6]^{3-} \) (\( d^5 \), weak field): 5 unpaired electrons (paramagnetic)
Step 2: Identify the number of unpaired electrons for each paramagnetic species.
The paramagnetic species have 4, 4, 2, 4, 1, and 5 unpaired electrons respectively.
Step 3: Count the number of paramagnetic species that share the same number of unpaired electrons.
One unpaired electron: \( [Fe(CN)_6]^{3-} \) (1 species)
Two unpaired electrons: \( [Mn(CN)_6]^{3-} \) (1 species)
Four unpaired electrons: \( [Co(C_2O_4)_3]^{3-} \), \( [MnCl_6]^{3-} \), \( [CoF_6]^{3-} \) (3 species)
Five unpaired electrons: \( [FeF_6]^{3-} \) (1 species)
The maximum number of paramagnetic species with the same number of unpaired electrons is 3 (all having 4 unpaired electrons).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: