Four, (1/3 points) There are 99 numbers $a_{1}, a_{2}, a_{3}, \cdots, a_{98}$, defined as: $S_{1} = a_{1}, S_{2} = a_{1} + a_{2}, \cdots, S_{k} = a_{1} + a_{2} + \cdots + a_{k} (1 \leqslant k \leqslant 99)$. There are 100 numbers $b_{1}, b_{2}, \cdots, b_{100}$, and $b_{1} = 1, b_{2} = 1 + a_{1}, b_{3} = 1 + a_{2}, \cdots, b_{k} = 1 + a_{k-2} (2 \leqslant k \leqslant 100)$, defined as: $S_{1}^{\prime} = b_{1}, S_{2}^{\prime} = b_{1} + b_{2}, \cdots, S_{k}^{\prime} = b_{1} + b_{2} + \cdots + b_{k} (1 \leqslant k \leqslant 100)$.
(1) Find: $S_{k}^{\prime} - S_{k-1} (2 \leqslant k \leqslant 100)$;
(2) If $A = \frac{S_{1} + S_{2} + \cdots + S_{90}}{99} = 50$, find the value of $B = \frac{S_{1}^{\prime} + S_{2}^{\prime} + \cdots + S_{100}^{\prime}}{100}$.
$\begin{array}{l}\text { Four, (1) } \because S_{k}^{\prime}=b_{1}+b_{2}+\cdots+b_{k} \\ ==?+\left(1+a_{1}\right)+\left(1+a_{2}\right)+\cdots+\left(1+a_{k-1}\right) \\ =k+\left(a_{0}+a_{2}+\cdots+a_{k-1}\right) \\ =k+S_{k-1} \text {. } \\ \therefore S^{\prime}{ }_{4}-S_{2-1}=k \text {. } \\ \text { (2) } \because A=\frac{S_{1}+S_{2}+\cdots+S_{99}}{99}=50 \text {, } \\ \therefore S_{1}+S_{2}+\cdots+S_{99}=50 \times 99 \text {. } \\ B=\frac{S_{2}{ }^{\prime}+S_{3}{ }^{\prime}+\cdots+S_{100}{ }^{\prime}}{100} \\ =\frac{1+\left(2+S_{1}\right)+\left(3+S_{2}\right)+\cdots+\left(100+S_{99}\right)}{100} \\ =\frac{100(100+1)}{2}+50 \times 99 \\ =\frac{101 \times 50+50 \times 99}{100} \\ =100 . \\\end{array}$