Sequence and series (8) reborn

Given that $(3-a_n)(6+a_{n-1}) = 18$ , then

$\begin{aligned} a_{n-1} & = 3-\frac {18}{6+a_{n-1}} = \frac {3a_{n-1}}{6+a_{n-1}} \\ \frac 1{a_n} & = \frac 2{a_{n-1}} + \frac 13 & \small \blue{\text{Let }b_k = \frac 1{a_k}} \\ b_n & = 2b_{n-1} + \frac 13 \\ b_0 & = \frac 1{a_0} = \frac 13 = \frac {2^1-1}3 \\ b_1 & = \frac 23 + \frac 13 = 1 = \frac {2^2-1}3 \\ b_2 & = 2 + \frac 13 = \frac 73 = \frac {2^3-1}3 \\ \implies b_n & = \frac {2^{n+1}-1}3 & \small \blue{\text{See proof below.}} \end{aligned}$

Therefore, $\displaystyle \sum_{n=0}^{100} b_n = \sum_{n=0}^{100} \frac {2^{n+1}-1}3 = \frac {2(2^{101}-1)-101}3 = \frac {2^{102}-103}3 \implies \alpha + \beta + \gamma = 102+103+ 3 = \boxed {208}$ .

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Proof: We can prove by induction that $b_n = \dfrac {2^{n+1}-1}3$ is true for $n \ge 0$ . The claim is true for $n=0$ as shown above. Assuming the claim is true for $n$ , then

$\begin{aligned} b_{n+1} & = 2b_n + \frac 13 = 2\times \frac {2^{n+1}-1}3 + \frac 13 = \frac {2^{n+2}-1}3 \end{aligned}$

The claim is also true for $n+1$ and hence true for all $n \ge 0$ .

The given recurrance relation yields $\dfrac{1}{a_i}=\dfrac{1}{3}+\dfrac{2}{a_{i-1}}$ .

So, $S=\dfrac{1}{a_0}+\dfrac{1}{a_1}+...+\dfrac{1}{a_{100}}=$

$\dfrac{101}{3}+2(S-\frac{1}{a_{100}})$ . And $\dfrac{1}{a_{100}}=\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{4}{3}+...+\dfrac{2^{100}}{3}=\dfrac{2^{101}-1}{3}$

$\implies S=\dfrac{2}{a_{100}}-\dfrac{101}{3}=\dfrac{2^{102}-103}{3}$ . Hence $α=102, β=103, \gamma=3$ , and $α+β+\gamma=102+103+3=\boxed {208}$ .