Sum of Digit Patterns and Natural Number Parameters

Let $T_r = 7\overbrace{5\dots 5}^{r}7$ denote the $(r+2)$-digit number. We can express $T_r$ mathematically as: $T_r = 7 \cdot 10^{r+1} + 5 \sum_{k=1}^{r} 10^k + 7$ Using the sum of a geometric progression, we get: $T_r = 7 \cdot 10^{r+1} + 5 \cdot \frac{10(10^r - 1)}{9} + 7 = \frac{680}{9} \cdot 10^r + \frac{13}{9}$

We are given the sum $S = \sum_{r=0}^{98} T_r$: $S = \sum_{r=0}^{98} \left( \frac{680}{9} \cdot 10^r + \frac{13}{9} \right) = \frac{680}{9} \left( \frac{10^{99} - 1}{9} \right) + \frac{13}{9} \times 99$ $S = \frac{680(10^{99} - 1) + 11583}{81} = \frac{680 \cdot 10^{99} + 10903}{81}$

Note that $T_{99} = 7\overbrace{5\dots 5}^{99}7 = \frac{680 \cdot 10^{99} + 13}{9}$, which implies: $680 \cdot 10^{99} = 9 T_{99} - 13$

Substituting this into the expression for $S$: $S = \frac{(9 T_{99} - 13) + 10903}{81} = \frac{9 T_{99} + 10890}{81} = \frac{T_{99} + 1210}{9}$

Comparing this with $S = \frac{T_{99} + m}{n}$, we obtain: $m = 1210, \quad n = 9$

Both $m$ and $n$ are natural numbers less than $3000$. Thus, the value of $m+n$ is: $m + n = 1210 + 9 = 1219$