Sum of annoying fractions

Simplifying all the fractions, we get $\frac{1+2}{3}+\frac{4+5}{6}+\frac{7+8}{9}+...+\frac{16+17}{18}$ .

Let $b=3$ .

Then, the sum is $\frac{2b-3}{b}+\frac{4b-3}{2b}+\frac{6b-3}{3b}+...+\frac{12b-3}{6b}$

Which simplifies to $2-\frac{3}{b}+2-\frac{3}{2b}+2-\frac{3}{3b}+...+2-\frac{3}{6b}$

and

$12-(\frac{3}{b}+\frac{3}{2b}+\frac{3}{3b}+...+\frac{3}{6b})$

We can factor out $\frac{3}{b}$ to get $12-\frac{3}{b}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6})$

Substituting $3$ back in the expression as $b$ , we get $12-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6})=12-2.45=9.55$ .

Or you could brute force it...