$\text{Finding Terms}$

Let's see first few terms of the sequence $a_2 = \frac{a_1 + 1}{2} = \frac{a_1 + 2^0 × 1}{2^1}$ $a_3 = \frac{a_2 + 2}{2} = \frac{a_1 + 2^0 × 1 + 2^1 × 2}{2^2}$ $a_4 = \frac{a_3 + 3}{2} = \frac{a_1 + 2^0 × 1 + 2^1 × 2 + 2^2 × 3}{2^3}$

We can get a pattern $a_r = \frac{a_{r - 1} + (r - 1)}{2} = \frac{a_1 + (2^0 × 1 + 2^1 × 2 \ldots 2^{r-3} × (r - 2) + 2^{r - 2} × (r - 1))}{2^{r - 1}}$ $\colorbox{#333333}{\color{#FFFFFF}{\boxed{a_r = \frac{a_1 + (2^{r - 1} × (r - 2) + 1)}{2^{r - 1}} = (r - 2) + \frac{a_1 + 1}{2^{r - 1}}}}}$

$\text{Finding Sum}$

Let's find the sum $\sum_{r = 1}^{r = \infty} [a_r - (r - 2)]= \sum_{r = 1}^{r = \infty} \frac{a_1 + 1}{2^{r - 1}}$ $\colorbox{#333333}{\color{#FFFFFF}{\boxed{\sum_{r = 1}^{r = \infty} [a_r - (r - 2)]= 2(a_1 + 1)}}}$ $\color{#3D99F6}{\boxed{\sum_{r = 1}^{r = \infty} [a_r - (r - 2)]= 2(898 + 1) = 1798}}$

A simpler solution

From the recursive relation given:

$\begin{aligned} a_r & = \frac {a_{r-1}+r - 1}2 \\ 2a_r & = a_{r-1} + t - 1 & \small \blue{\text{Replace }r \text{ with }r+1.} \\ 2a_{r+1} & = a_r + r \\ 2a_{r+1} - 2r + 2 & = a_r - r +2 \\ 2a_{r+1} -2(r+1) + 4 & = a_r - r + 2 & \small \blue{\text{Let }b_r = a_r - r + 2} \\ 2b_{r+1} & = b_r \\ \implies b_{r+1} & = \frac {b_r}2 \\ \implies b_r & = \frac {b_1}{2^{r-1}} \end{aligned}$

Therefore, $\displaystyle \sum_{r=1}^\infty (a_r - r+2) = \sum_{r=1}^\infty b_r = \sum_{r=0}^\infty \frac {b_1}{2^r} = \frac {b_1}{1-\frac 12} = 2b_1 = 2(a_1 - 1 + 2) = \boxed{1798}$