A sequence $T_n$ is given by $T_n = ext{ln}(n^a) + bn^2 + c$ for $n ext{ } oldsymbol{ m ackslash in} ext{ } oldsymbol{ m ackslash Z}^+$. The first three terms of the sequence are $9$, $17.306$ and $31.901$.

1. Formulate a system of linear equations using the first three terms of the sequence.

Given $T_n = \ln(n^a) + bn^2 + c$, we substitute $n=1, 2, 3$ to form equations based on the given terms.

For $n = 1$:
$T_1 = \ln(1^a) + b(1)^2 + c = 9$
$0 + b + c = 9$ --- (1)

For $n = 2$:
$T_2 = \ln(2^a) + b(2)^2 + c = 17.306$
$a\ln 2 + 4b + c = 17.306$ --- (2)

For $n = 3$:
$T_3 = \ln(3^a) + b(3)^2 + c = 31.901$
$a\ln 3 + 9b + c = 31.901$ --- (3)

2. Solve the system of equations simultaneously to find the values of $a$, $b$, and $c$.

From (1), $c = 9 - b$. Substitute this into (2) and (3):
$a\ln 2 + 4b + (9 - b) = 17.306 \implies a\ln 2 + 3b = 8.306$ --- (4)
$a\ln 3 + 9b + (9 - b) = 31.901 \implies a\ln 3 + 8b = 22.901$ --- (5)

Multiply (4) by $8$ and (5) by $3$ to align coefficients for $b$:
$8a\ln 2 + 24b = 66.448$
$3a\ln 3 + 24b = 68.703$

Subtracting the first aligned equation from the second to eliminate $b$:
$a(3\ln 3 - 8\ln 2) = 2.255$
$a(-2.2493) \approx 2.255 \implies a \approx -1.0025$

Substitute $a$ back into (4) to find $b$:
$-1.0025\ln 2 + 3b = 8.306 \implies 3b \approx 9.0009 \implies b \approx 3.0003$

Substitute $b$ to find $c$:
$c = 9 - b \approx 9 - 3.0003 = 5.9997$

3. Round the calculated values to the nearest integer as required.

Rounding the values of $a$, $b$, and $c$ to the nearest integers:
$a = -1$
$b = 3$
$c = 6$

1. Determine the values of $a$, $b$, and $c$ from the given first three terms.

$T_1 = \ln(1^a) + b(1)^2 + c = 9 \implies b + c = 9$
$T_2 = a\ln(2) + 4b + c = 17.306$
$T_3 = a\ln(3) + 9b + c = 31.901$
Subtracting the first equation from the second and third gives:
$a\ln 2 + 3b = 8.306$
$a\ln 3 + 8b = 22.901$
Solving these simultaneously yields $a \approx -1.002$, $b \approx 3.000$, and $c \approx 6.000$. Rounding to the nearest integers, we get $a = -1$, $b = 3$, $c = 6$.

2. Evaluate the sum of the first 100 terms of the sequence.

Substitute $a, b, c$ into the sequence formula:
$T_n = \ln(n^{-1}) + 3n^2 + 6 = -\ln n + 3n^2 + 6$
The sum of the first 100 terms is:
$S_{100} = \sum_{n=1}^{100} (-\ln n + 3n^2 + 6)$
We can evaluate this by applying summation formulas and evaluating with a graphing calculator:
$S_{100} = -\sum_{n=1}^{100} \ln n + 3\sum_{n=1}^{100} n^2 + \sum_{n=1}^{100} 6$
$S_{100} = -\ln(100!) + 3\left(\frac{100 \times 101 \times 201}{6}\right) + 600$
$S_{100} = 1015650 - \ln(100!) \approx 1015286.26$
To 3 significant figures, the sum is $1.02 \times 10^6$.