jmc

Taking the first product, we have $\left(1-\frac{1}{2}\right)=\frac{1}{2}$ $\frac{1}{2}\times 2=1$ Looking at the second, we get $\left(1-\frac{1}{3}\right)=\frac{2}{3}$ $\frac{2}{3}\times 3=2$ We seem to be going up by $1$. Just to check, $1-\frac{1}{n}=\frac{n-1}{n}$ $\frac{n-1}{n}\times n=n-1$ Now that we have discovered the pattern, we have to find the last term. $1-\frac{1}{10}=\frac{9}{10}$ $\frac{9}{10}\times 10=9$ The sum of all numbers from $1$ to $9$ is $\frac{9\cdot 10}{2}=45=\boxed{45}$