$2\left(1-\dfrac{1}{2}\right) + 3\left(1-\dfrac{1}{3}\right) + 4\left(1-\dfrac{1}{4}\right) + \cdots + 10\left(1-\dfrac{1}{10}\right)=$
$\text{(A)}\ 45 \qquad \text{(B)}\ 49 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 54 \qquad \text{(E)}\ 55$
Taking the first product, we have
$\left(1-\frac{1}{2}\right)=\frac{1}{2}$
$\frac{1}{2}\times2=1$
Looking at the second, we get
$\left(1-\frac{1}{3}\right)=\frac{2}{3}$
$\frac{2}{3}\times3=2$
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.
$1-\frac{1}{10}=\frac{9}{10}$
$\frac{9}{10}\times10=9$
The sum of all numbers from $1$ to $9$ is
$\frac{9\cdot10}{2}=45=\boxed{A}$