Tower of 2 Summation

Let $S_n = 1 + \displaystyle \sum_{r=0}^n \frac {2^{2^r+r-1}}{1+2^{2^r}}$ . First few $n$ s show that $S_n = \dfrac {2^{a_n}}{2^{b_n}-1}$ , where $a_n = 2^{n+1}+n$ and $b_n = 2^{n+1}$ . Let us prove the induction that the claim is true for all $n \ge 0$ .

Proof: For $n=0$ , $S_0 = 1 + \dfrac {2^{2^0+0-1}}{1+2^{2^0}} = 1 + \dfrac 13 = \dfrac 43 = \dfrac {2^2}{2^2-1}$ . Note that $a_0 = 2^{0+1} + 0 = 2$ and $b_n = 2^{0+1} = 2$ . Therefore, the claim is true for $n=0$ . Now assuming that the claim is true for $n$ , then:

$\begin{aligned} S_{n+1} & = 1 + \sum_{r=0}^{n+1} \frac {2^{2^r+r-1}}{1+2^{2^r}} \\ & = S_n + \frac {2^{2^{n+1}+n}}{1+2^{2^{n+1}}} \\ & = \frac {2^{2^{n+1}+n}}{2^{2^{n+1}}-1} + \frac {2^{2^{n+1}+n}}{2^{2^{n+1}}+1} \\ & = 2^{2^{n+1}+n} \left(\frac 1{2^{2^{n+1}}-1} + \frac 1{2^{2^{n+1}}+1} \right) \\ & = 2^{2^{n+1}+n} \left(\frac {2^{2^{n+1}}+1+2^{2^{n+1}}-1}{\left(2^{2^{n+1}}-1\right)\left(2^{2^{n+1}}+1\right)} \right) \\ & = 2^{2^{n+1}+n} \left(\frac {2\times 2^{2^{n+1}}}{2^{2^{n+1}\times 2}-1^2} \right) \\ & = \frac {2^{2^{n+2}+n+1}}{2^{2^{n+2}}-1} \\ & = \frac {2^{a_{n+1}}}{2^{b_{n+1}}-1} \end{aligned}$

Therefore, the claim is also true for $n+1$ implying it is true for all $n\ge 0$ .

Therefore, $\alpha = a_{10} = 2^{11} + 10 = 2058$ , $\beta = b_{10} = 2^{11} = 2048$ and $\alpha + \beta + \alpha\beta = \boxed{4218890}$ .