$7 \cdot 96$ If $F(n+1)=\frac{2 F(n)+1}{2}, n=1,2, \cdots$, and $F(1)=2$, then $F(101)$ equals
(A) 49.
(B) 50.
(C) 51.
(D) 52.
(E) 53.
(17th American High School Mathematics Examination, 1966)
[Solution] Given $F(n+1)=F(n)+\frac{1}{2}, n=1,2, \cdots$. This indicates an arithmetic sequence with a common difference of $\frac{1}{2}$. And $F(1)=2$, which means the first term is 2.
Therefore, $F(n)=2+\frac{n-1}{2}$.
Thus, $F(101)=2+\frac{101-1}{2}=52$.
Hence, the answer is $(D)$.