Use linearity and the following formulas to rewrite the sum. \[ \begin{aligned} \sum_{j=1}^{N} j & =\frac{N(N+1)}{2} \\ \sum_{j=1}^{N} j^{2} & =\frac{N(N+1)(2 N+1)}{6} \end{aligned} \] Evaluate the sum. \[ \sum_{m=1}^{20}\left(3+\frac{3 m}{2}\right)^{2} \] (Use decimal notation. Give your answer to one decimal place.) \[ \sum_{m=1}^{20}\left(3+\frac{3 m}{2}\right)^{2}= \]

Step 1 ：Given the sum \(\sum_{m=1}^{20}\left(3+\frac{3 m}{2}\right)^{2}\)

Step 2 ：Expand the square in the sum using the formula for the square of a binomial, \((a+b)^2 = a^2 + 2ab + b^2\), where \(a = 3\) and \(b = \frac{3m}{2}\)

Step 3 ：This gives us \(\sum_{m=1}^{20} (9 + 9m + \frac{9m^2}{4})\)

Step 4 ：Separate this into three different sums: \(\sum_{m=1}^{20} 9 + \sum_{m=1}^{20} 9m + \sum_{m=1}^{20} \frac{9m^2}{4}\)

Step 5 ：Use the provided formulas to evaluate each sum: \(\sum_{j=1}^{N} j =\frac{N(N+1)}{2}\) and \(\sum_{j=1}^{N} j^{2} =\frac{N(N+1)(2 N+1)}{6}\), where \(N = 20\)

Step 6 ：Calculate the first sum: \(\sum_{m=1}^{20} 9 = 20 \times 9 = 180\)

Step 7 ：Calculate the second sum: \(\sum_{m=1}^{20} 9m = 9 \times \frac{20(20+1)}{2} = 1890.0\)

Step 8 ：Calculate the third sum: \(\sum_{m=1}^{20} \frac{9m^2}{4} = \frac{9}{4} \times \frac{20(20+1)(2 \times 20+1)}{6} = 6457.5\)

Step 9 ：Add the three sums together to get the total sum: \(180 + 1890.0 + 6457.5 = 8527.5\)

Step 10 ：Final Answer: \(\boxed{8527.5}\)