Remainders Ahead!

Relevant wiki: Remainder Factor Theorem - Intermediate

Let

$\begin{aligned} f(x) & = g(x) q(x) + \color{#3D99F6} r(x) & \small \color{#3D99F6} \text{where }r(x) \text{ is the remainder polynomial.} \\ x^{101} & = (x^2+3x+2)q(x) + \color{#3D99F6} ax+b & \small \color{#3D99F6} \text{Note that }r(x) \text{ has a maximum degree of 1.} \\ x^{101} & = (x+1)(x+2)q(x) + ax+b & \small \color{#3D99F6} \text{Putting }x=-1 \\ -1 & = -a+b \quad ...(1) & \small \color{#3D99F6} \text{Putting }x=-2 \\ -2^{101} & = -2a+b \quad ...(2) \end{aligned}$

$\begin{aligned} (1)-(2): \quad a & = 2^{101}-1 \\ (1): \quad b & = a-1 = 2^{101}-2 \end{aligned}$

Therefore,

$\begin{aligned} \frac {2^{50}b}{a+1+2^{51}}+1 & = \frac {2^{50}(2^{101}-2)}{2^{101}-1+1+2^{51}}+1 \\ & = \frac {2^{50}(2^{101}-2) +2^{101}+2^{51} }{2^{101}+2^{51}} \\ & = \frac {2^{50}(2^{101}+2^{51})}{2^{101}+2^{51}} \\ & = \boxed{2^{50}} \end{aligned}$