It is given that \(\dfrac{r-2}{2^r}=\dfrac{Ar}{2^r}+\dfrac{B(r-1)}{2^{r-1}}\), where \(A\) and \(B\) are real constants.
Find the value of \(A\) and of \(B\).[2]
Hence show that \(\displaystyle\sum_{r=1}^n\dfrac{r-2}{2^r}=-\dfrac n{2^n}\).
Given that \(2^n\) is much larger than \(n\) for large values of \(n\), state the value of \(\displaystyle\sum_{r=1}^\infty\dfrac{r-2}{2^r}\).[4]
Hence find the value of \(\displaystyle\sum_{r=1}^\infty\dfrac{r-1}{2^r}\).[3]